Carol ate 2/5 of the cake, Dima ate 3/8 of the REMAINING cake, mom ate the rest.
If it was a POUND cake, how many OZ did MOM eat? PLEASE HELP ASAP, GIVING BRAINLIEST.

Answers

Answer 1
she would've eaten half of the cake, or 8 oz.
Answer 2

Her mum ate 31/40 of the cake

Given the weight of the cake to 1 pound

If Carol ate 2/5 of the cake, the remaining fraction of the cake will be 1 - 2/5 = 3/5

If Dima ate 3/8 of the REMAINING cake, then the fraction of cake Dima ate will be:

Dima = 3/8 * 3/5

Dima = 9/40

Fraction of the cake that her mum ate = 1 - 9/40 = 31/40

Hence her mum ate 31/40 of the cake

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Related Questions

Jade spent 37.60 on groceries. 4/5 of that total was spent on vegetables. How much was spent on other items

Answers

total amount=4/5(total amount)+unknown amount
37.6=4/5(37.6)+x
(1/5)(37.6)=x
x=37.6/5

What is the value of the expression? 56−(18÷34) as a fraction in simplest form

Answers

The value of the expression in simplest form is: 55 8/17.

Hope this helps! :D

~PutarPotato

The value of the expression [tex]\( 56 - \left( \frac{18}{34} \right) \)[/tex] as a fraction in simplest form is [tex]\( \frac{943}{17} \)[/tex].

To solve the expression, we first perform the division within the parentheses:

[tex]\[ \frac{18}{34} \][/tex]

This fraction can be simplified by dividing both the numerator and the denominator by their greatest common divisor, which is 2:

[tex]\[ \frac{18 \div 2}{34 \div 2} = \frac{9}{17} \][/tex]

Now we have:

[tex]\[ 56 - \frac{9}{17} \][/tex]

We need to express the whole number as a fraction with the same denominator as the fraction we are subtracting.

[tex]\[ \frac{56 \times 17}{17} - \frac{9}{17} \][/tex]

[tex]\[ \frac{952}{17} - \frac{9}{17} \][/tex]

[tex]\[ \frac{952 - 9}{17} \][/tex]

[tex]\[ \frac{943}{17} \][/tex]

solve the equation for x.

ax+bx=c

Answers

ax+bx = c
abx = c
abx/ab = c/ab
x = c/ab

Answer: x = c/a + b

Step-by-step explanation: I have put the variable x in purple in the image that I have made so you can see it better.

Notice that the variable we are solving for in

this problem appears in 2 separate terms.

In this situation, we must factor the x out

of both terms to get x(a + b) = c.

Think about it as the reverse of the distributive property.

In other words, if we take our second step and distribute the x

through the parentheses, we end up back at our first step.

The beauty of factoring is that the two x's in this problem

have now been combined into one x and we can get x

by itself by dividing both sides by a + b.

This gives us x = c/a + b.

Help please :) ! The dashed triangle is the image of the pre-image solid triangle. What is the scale factor used to create the dilation? Enter your answer in the box.

Answers

Let

x---------> measure of the base of the triangle of the image

y---------> measure of the base of the triangle of the pre-image

sf-------> the scale factor

we know that

the scale factor is equal to

[tex]sf=\frac{x}{y}[/tex]

in this problem we have

[tex]x=6\ units\\y=3\ units[/tex]

Find the scale factor

substitute the values of x and y in the formula above

[tex]sf=\frac{6}{3}=2[/tex]

therefore

the answer is

the scale factor is equal to [tex]2[/tex]


Discuss the differences between parametric and non-parametric tests?

Answers

Nonparametric tests are also called distribution-free tests because they don't assume that your data follow a specific distribution. You may have heard that you should use nonparametric tests when your data don't meet the assumptions of the parametric test, especially the assumption about normally distributed data.

We are conducting a test of the hypotheses h0: p = 0.2 ha: p ≠ 0.2 we find a test statistic of z = -1.11. what is the corresponding p-value? give your answer as a proportion between 0 and 1 to 4 decimal places.

Answers

we have to find the p value from the z table and multiply by 2 for two tailed hypothesis at 5% level of significance. The P-Value is 0.266999.

12.917 rounded to the nearest ones

Answers

I am pretty sure the answer is 13
To round to the nearest ones, round to the nearest digit for the number 2. Since 9 rounds up, then the answer is 13.

Amanda, Bryce, and Corey enter a race in which they have to run, swim, and cycle over a marked course. Their average speeds are given in the table. Corey finishes first with a total time of 2 hr 32 min. Amanda comes in second with a time of 3 hr 33 min. Bryce finishes last with a time of 4 hr 14 min. Find the distance (in mi) for each part of the race.

Answers

How far is the course? Can you provide the table the question say is included?

Computer upgrades have a nominal time of 80 minutes. samples of five observations each have been taken, and the results are as listed. sample 1 2 3 4 5 6 79.2 80.5 79.6 78.9 80.5 79.7 78.8 78.7 79.6 79.4 79.6 80.6 80.0 81.0 80.4 79.7 80.4 80.5 78.4 80.4 80.3 79.4 80.8 80.0 81.0 80.1 80.8 80.6 78.8 81.1 factors for three-sigma control limits for formula179.mml and r charts factors for r charts number of observations in subgroup, n factor for formula180.mml chart, a2 lower control limit, d3 upper control limit, d4 2 1.88 0 3.27 3 1.02 0 2.57 4 0.73 0 2.28 5 0.58 0 2.11 6 0.48 0 2.00 7 0.42 0.08 1.92 8 0.37 0.14 1.86 9 0.34 0.18 1.82 10 0.31 0.22 1.78 11 0.29 0.26 1.74 12 0.27 0.28 1.72 13 0.25 0.31 1.69 14 0.24 0.33 1.67 15 0.22 0.35 1.65 16 0.21 0.36 1.64 17 0.20 0.38 1.62 18 0.19 0.39 1.61 19 0.19 0.40 1.60 20 0.18 0.41 1.59
a. using factors from above table, determine upper and lower control limits for mean and range charts. (round your intermediate calculations and final answers to 2 decimal places. leave no cells blank - be certain to enter "0" wherever required.) mean chart range chart ucl 81.06 4.01 lcl 78.86 0.00 g

Answers

Final answer:

The UCL and LCL for the mean and range charts are calculated based on the average and range of the data sets. The table provided in the question provides respective factors for various subgroup sizes. For n=5, the factors are A2=0.58, D3=0 and D4=2.11. Using these we compute UCL and LCL for both charts. Round off to two decimal places and fill '0' in any empty cells.

Explanation:

To find the Upper Control Limit (UCL) and Lower Control Limit (LCL) for the mean and range charts, we have to use the r and formula179.mml charts that are provided in your question. We will take the observations in groups of five as mentioned in the question.

Firstly, calculate the average (X-bar) for each of these groups and find the overall average (M) of these five averages.Please find the range (R) for each subgroup and then the average of these ranges (meanR).Referring to the table given in your question, for n=5, the factors are A2=0.58, D3=0 and D4=2.11.The UCL and LCL for the mean chart can be calculated using UCL=M+A2*meanR and LCL=M-A2*meanR, whereas, for the range chart, they are UCL=D4*meanR and LCL=D3*meanR

Using these steps, you can find the upper and lower control limits for both the mean and range charts. Always remember to round off your answers to two decimal places and also, to avoid leaving any cells blank in the excel sheet. If required, fill in  '0' in those empty cells. This way, meaningful interpretation of the control charts can be made.

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The upper and lower control limits for the mean chart are 81.06 and 78.86, respectively, while the upper control limit for the range chart is 8.45 and the lower control limit is 0.

To determine the upper and lower control limits for the mean and range charts, we can use the factors provided in the table.

For the mean chart:

- The upper control limit (UCL) is calculated by adding the product of the factor for the mean chart (A2) and the average range (R-bar) to the overall average (X-double-bar). In this case, the A2 factor is 0.73 and the R-bar is 4.01, so the UCL for the mean chart is: 81.06 (rounded to 2 decimal places).

- The lower control limit (LCL) is calculated by subtracting the product of the factor for the mean chart (A2) and the average range (R-bar) from the overall average (X-double-bar). Using the same values, the LCL for the mean chart is: 78.86 (rounded to 2 decimal places).

For the range chart:

- The UCL for the range chart is calculated by multiplying the factor for the range chart (D4) by the average range (R-bar). In this case, the D4 factor is 2.11 and the R-bar is 4.01, so the UCL for the range chart is: 8.45 (rounded to 2 decimal places).

- The LCL for the range chart is always 0 since range values cannot be negative.

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Suppose 4 mg of a drug is injected into a person's bloodstream. As the drug is metabolized, the quantity diminishes at the continuous rate of 3% per hour?

Answers

Let Q(t) = the mass (mg) remaining after t hours.
We are told that Q diminishes by 3% every hour.

When t = 0, Q = 4 mg
When t = 1,  Q = 4*0.97 mg
When t = 2, Q = 4*(0.97)² mg

By induction,
[tex]Q(t) = 4 (0.97)^{t} \, mg[/tex]

Q'(t) = 0.97Q
Therefore the rate of decrease is 3% per hour.

The person receives an additional dosage when Q falls to 0.50 mg. This happens when
[tex]4(0.97)^{t}=0.5 \\0.97^{t} = 0.125 \\ t \,ln(0.97) = ln(0.125) \\ t = \frac{ln(0.125)}{ln(0.97)} =68.27 \, hrs[/tex]

After the second injection, the mass is now 4.5 mg. Therefore
[tex]Q(t)=4.5(0.97)^{t}[/tex]

When the mass again reaches 0.50 mg, then
[tex]4.5(0.97)^{t}=0.5 \\ 0.97^{t}=0.1111 \\ t \, ln(0.97) = ln(0.1111) \\ t = 72.14 \, hrs[/tex]

Answers: 
(a) [tex]Q(t) = 4(0.97)^{t}[/tex]
(b) 3% per hour
(c) 68.3 hours
(d) 72.1 hours

Write this in Interval Notation:

4x - 6 ≥ 6x - 20

Answers

4x - 6 ≥ 6x - 20 = 6x-20,∞ 

the sum of partial products is equal to the final product?

Answers

the sum of partial product is not always equal to a final product

You have a photo that measures 1 in. wide by 3 in. tall. If you want to enlarge the photo so that the height is 18 in., what will the width of the photo be?

A. 98 in.
B. 4 in.
C. 6 in.
D. 108 in.

Answers

C. 6in is the answer
18/3= 6

6(1)= 6

The answer is C.

Hope this helps!

The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds. the probability of a player weighing more than 241.25 pounds is

Answers

Final answer:

To find the probability of a football player weighing more than 241.25 pounds, calculate the z-score and use the standard normal distribution table.

Explanation:

To find the probability of a football player weighing more than 241.25 pounds, we need to calculate the z-score and then use the standard normal distribution table.

The formula for calculating the z-score is: z = (x - μ) / σ where x is the player's weight, μ is the mean weight, and σ is the standard deviation.

Plugging in the values, we get: z = (241.25 - 200) / 25 = 1.65

Using the standard normal distribution table, we can find that the probability of a value being greater than 1.65 is approximately 0.0495.

Solve for Z. Z−4/9−1/3=5/9 Enter your answer in the box. Z=

NEED HELP FAST WILL GIVE BRAINLIEST 15 POINTS

Answers

z-4/9-1/3=5/9
z -4/9 + 4/9 - 1/3 = 5/9 + 4/9 = 1
z - 1/3 + 1/3 = 1 + 1/3
z= 1 1/3

Answer:

The answer correct answer is, 4/3

Step-by-step explanation:

I am reveiwing this quiz right now!

You have 3/8 of a cake. You divide the remaining cake into six equal pieces. What fraction of the original cake is each piece?

Answers

Just divide 3/8 (of a cake) by 6.  That comes out to    3
                                                                                   ------ = 1/16 of the original
                                                                                    8(6)      cake.

Which equation is equivalent to 1 5/6 - 2/6 = 1 3/6

Answers

11/6-1/3 = 9/6 should be equivalent

There are 56 trees in an apple orchard. They are arranged in equal rows. There are 8 trees in each row. How many rows of apple trees are there? Which equation can be used to solve this problem?

Answers

Final answer:

In this case, there are 56 trees in the orchard and 8 trees in each row, so there are 7 rows of apple trees in the orchard.

Explanation:

To find the number of rows of apple trees in the orchard, we need to divide the total number of trees by the number of trees in each row.

In this case, there are 56 trees in the orchard and 8 trees in each row.

We can use the equation:

Number of rows = Total number of trees / Number of trees in each row

Plugging in the values:

Number of rows = 56 trees / 8 trees = 7 rows

Therefore, there are 7 rows of apple trees in the orchard.

How do I multiply 8.354 by 11.81?

Answers

well you can do it manually or use a calculator.

Answer is  98.66074

all quadrilaterals are polygons

Answers

True.
Polygons are figures with straight edges, and any number of sides/vertices.
Quadrilaterals are figures with straight edges, with 4 sides and 4 vertices.
true (sorry for this but my text cant be less than 20 characters)

a number is less than 200 and greater than 100 the ones digit is 5 less than 10.the tens digit is 2 more than the ones digit.what is the number

Answers

It's between 101 and 199.

5 less than 10 is 5.
2 more than 5 is 7

My guess is 157

The correct number is between 100 and 200 would be 175.

Used the concept of number system which states that,

A writing system used to express numbers is known as a number system. It is the mathematical notation used to consistently express the numbers in a particular set using digits or other symbols.

Given that,

A number is less than 200 and greater than 100.

Here, the one's digit is 5 less than 10, and the tens digit is 2 more than the one digit.

Since the one's digit is 5 less than 10.

So, the number in one's place is,

10 - 5 = 5

And, the tens digit is 2 more than the one digit.

5 + 2 = 7

Since the number is between 100 and 200.

So, the possible value of a number with the value in one's place 5 and the value in tens place 7 would be,

175

Clearly, the number 175 is between 100 and 200.

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a box of cereal contains 23.4 ounces. It cost $5.49. What is the cost,to the nearest cent, of the cereal per ounce.

Answers

Do 5.49 / 23.4
This equals $0.23

The answer is $0.23

One root of is x = 6. What are all the factors of the function? Use the Remainder Theorem.

Answers

3 and 2 should be the only roots

Answer:

C. (x - 6)(x + 4)(x - 2)

Step-by-step explanation:

i just took the test goodluck

Jim Abbott purchased a $58,000 recreational vehicle (RV) with a 45 percent markup on selling price. a. What was the amount of the dealer’s markup? Markup amount $ b. What was the dealer’s original cost? Original cost $

Answers

Final answer:

The dealer's markup amount is $26,100 and the original cost of the RV is $31,900.

Explanation:

To determine the dealer's markup, we need to find 45% of the selling price of the RV. We can set up the following equation:

Markup amount = Selling price × Markup percentage

Let's plug in the given values: $58,000 × 45% = $26,100

The dealer's original cost can be found by subtracting the markup amount from the selling price:

Original cost = Selling price - Markup amount

Let's calculate it: $58,000 - $26,100 = $31,900

The table below shows the typical hours worked by employees at a company. A salaried employee makes $67,000 per year. Hourly employees get paid $25 per hour, but get $37.50 per hour for each hour over 40 hours. Sun. Mon. Tues. Wed. Thurs. Fri. Sat. 0 10 8 8 7 6.5 4.5 Which of the payment options would you recommend to a new employee? a. Either one. Hourly and salaried employees earn the same amount per week. b. Hourly pay. Hourly employees make more per week than salaried employees. c. Salaried pay. Salaried employees make more per week than hourly employees. d. There is not enough information given to compare weekly earnings.

Answers

First, we must calculate the weekly pay of an employee that is paid a fixed amount. Given that there are 52 weeks in a year, the weekly pay for a regularly paid employee is: 67,000 / 52 = $1,288.46 Now, we calculate the number of hours an employee that is paid hourly works per week: 0 + 10 + 8 + 8 + 7 + 6.5 + 4.5 = 44 So this employee is paid: 25 x 40 + 37.5 x 4 = $1,150 Therefore, it is recommended that a new employee goes for the salaried pay since the weekly earnings are greater in this option. The answer is C.

Answer:

C should be the correct option

Step-by-step explanation:

Tell me if im wrong :)

Your mid-morning break begins at 10:30 A.M. and lasts for 15 minutes. You work 2 hours and 50 minutes until your lunch break. What time does your lunch break begin?

Answers

1. You start by adding 15 minutes to 10:30
because the break takes 15 minutes Which equals 10:45.

2. You then add another 2 hours and 50 minutes of work until your lunch break which makes it 1:35.

3. Your lunch break starts at 1:35

Final answer:

The lunch break begins at 1:20 P.M. after working for 2 hours and 50 minutes from the mid-morning break at 10:30 A.M.

Explanation:

The mid-morning break begins at 10:30 A.M. and lasts for 15 minutes.

You work 2 hours and 50 minutes from 10:30 A.M.

To find the start time of your lunch break, add 2 hours and 50 minutes to 10:30 A.M.

Lunch break begins at 1:20 P.M.

is 1 1/3 a unit rate

Answers

No it is not a unit rate unless 11/3 is your numerator for a quantity of one. For example, "he finishes 11/3 of his homework every hour"
No, a unit rate always has a denominator of 1. For example, 5 miles an hour is 5/1.

To graph the inequality t > 5, you would put an open circle on 5 and shade to the right. true or false

Answers

true, open since t>5 and to the right since all values greater than 5 satisy
Since the sign is simply "greater than", not "greater than or equal to", you do use an open circle.
All numbers greater than 5 are to the right of 5, so you shade to the right of 5.
The statement is true.

Jacob used 1/7 of a liter of water to fill 1/9 of the fish aquarium. How many liters r needed to fill the aquarium?

Answers

Since 1/7 L of water fills 1/9 of the aquarium, 9 times 1/7 will fill the aquarium:

9(1/7) = 9/7 = 1 2/7 L of water will fill the aquarium.

we know that 1/7 of a liter filled up 1/9 of the aquarium, well, a "whole" will be 9/9.

how many liters will it be to fill up the 9/9 of the aquarium?

[tex]\bf \begin{array}{ccll} \stackrel{liters}{water}&\stackrel{fish}{aquarium}\\ \text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\ \frac{1}{7}&\frac{1}{9}\\\\ w&\frac{9}{9} \end{array}\implies \cfrac{\frac{1}{7}}{w}=\cfrac{\frac{1}{9}}{\frac{9}{9}}\implies \cfrac{\frac{1}{7}}{\frac{w}{1}}=\cfrac{\frac{1}{9}}{\frac{9}{9}}\implies \cfrac{1}{7}\cdot \cfrac{1}{w}=\cfrac{1}{9}\cdot \cfrac{9}{9} \\\\\\ \cfrac{1}{7w}=\cfrac{1}{9}\implies 9=7w\implies \cfrac{9}{7}=w[/tex]

Find the least common multiple (LCM) of 54 and 72. A) 9 B) 27 C) 108 D) 216

Answers

the lowest common multiple is 216 (D)
LCM (54,72):
54, 108, 162, 216...
72, 144, 216...
Answer: 
D) 216
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At friendly meetings, and when the wine was to his taste, something eminently human beaconed from his eye; something indeed which never found its way into his talk, but which spoke not only in these silent symbols of the after-dinner face, but more often and loudly in the acts of his life. He was austere with himself; drank gin when he was alone, to mortify a taste for vintages; and though he enjoyed the theatre, had not crossed the doors of one for twenty years. But he had an approved tolerance for others; sometimes wondering, almost with envy, at the high pressure of spirits involved in their misdeeds; and in any extremity inclined to help rather than to reprove."I incline to, Cain's heresy*," he used to say. "I let my brother go to the devil in his quaintly 'own way.'" In this character, it was frequently his fortune to be the last reputable acquaintance and the last good influence in the lives of down-going men. And to such as these, so long as they came about his chambers, he never marked a shade of change in his demeanour.No doubt the feat was easy to Mr. Utterson; for he was undemonstrative at the best, and even his friendship seemed to be founded in a similar catholicity of good-nature. It is the mark of a modest man to accept his friendly circle ready-made from the hands of opportunity; and that was the lawyer's way. His friends were those of his own blood or those whom he had known the longest; his affections, like ivy, were the growth of time, they implied no aptness in the object. Hence, no doubt, the bond that united him to Mr. Richard Enfield, his distant kinsman, the well-known man about town. It was a nut to crack for many, what these two could see in each other, or what subject they could find in common. 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