in a simple random sample of 48 men, the mean wrist circumference of the men is 19.3 centimeters with a standard deviation of 1.2 centimeters. Round your answer to the nearest hundredth.

Which interval is the 95% confidence interval for the mean wrist circumference for all men?

18.99,19.61
18.96, 19.64
12.85, 19.75
19.14, 19.46

Answer:  The answer is (C) 10 to 25 centimetres.

Step-by-step explanation: We are to select the correct range by which the sea level has risen about over the last 100 years.

Scientists estimates that because of warming of the ocean and increased melting of land-based ice such as glaciers and ice sheets, the level of sea has risen from 10 to 25 centimetres over the last 100 years.

Thus, (C) is the correct option.

The depth of the water is increasing at a rate of approximately is 0.14 ft/min

To determine how fast the depth of the water is increasing when the water is 13 feet deep, we need to relate the volume of the water in the conical tank to its depth. We can use related rates and the geometry of the cone.

Given:

- The radius of the tank at the top R = 10 feet

- The height of the tank H = 27 feet

- The rate of water flow into the tank [tex](\(\frac{dV}{dt}\))[/tex] = 10 ft[tex]\(^3\)/min[/tex]

- The depth of the water h = 13 feet

First, let's find the relationship between the radius r of the water's surface at depth h.

Since the water forms a smaller cone similar to the tank, we can use the concept of similar triangles:

[tex]\[\frac{r}{h} = \frac{R}{H} \implies \frac{r}{h} = \frac{10}{27} \implies r = \frac{10}{27}h\][/tex]

Next, we use the volume formula for a cone:

[tex]\[V = \frac{1}{3} \pi r^2 h\][/tex]

Substitute [tex]\(r = \frac{10}{27}h\):[/tex]

[tex]\[V = \frac{1}{3} \pi \left( \frac{10}{27}h \right)^2 h = \frac{1}{3} \pi \frac{100}{729} h^3 = \frac{100\pi}{2187} h^3\][/tex]

Differentiate both sides with respect to t:

[tex]\[\frac{dV}{dt} = \frac{100\pi}{2187} \cdot 3h^2 \frac{dh}{dt}\][/tex]

Simplify:

[tex]\[\frac{dV}{dt} = \frac{300\pi}{2187} h^2 \frac{dh}{dt}\][/tex]

Given [tex]\(\frac{dV}{dt} = 10 \) ft\(^3\)/min[/tex], and [tex]\(h = 13\) ft:[/tex]

[tex]\[10 = \frac{300\pi}{2187} \cdot (13)^2 \cdot \frac{dh}{dt}\][/tex]

Solve for [tex]\(\frac{dh}{dt}\):[/tex]

[tex]\[10 = \frac{300\pi}{2187} \cdot 169 \cdot \frac{dh}{dt}\][/tex]

[tex]\[10 = \frac{50700\pi}{2187} \cdot \frac{dh}{dt}\][/tex]

[tex]\[\frac{dh}{dt} = \frac{10 \cdot 2187}{50700\pi}\][/tex]

[tex]\[\frac{dh}{dt} = \frac{21870}{50700\pi}\][/tex]

[tex]\[\frac{dh}{dt} \approx \frac{21870}{159252} \approx \frac{1}{7.29} \approx 0.137 \text{ ft/min}\][/tex]

Therefore, the depth of the water is increasing at a rate of approximately: [tex]\[\boxed{0.14 \text{ ft/min}}\][/tex]

1814400 ways can 8 students be assigned a seat in a classroom if there are 10 seats in a row.

A permutation is a mathematical technique that determines the number of possible arrangements in a set when the order of the arrangement.

[tex]nP_{r} =\frac{n!}{(n-r)!}[/tex]

n=Total number of objects

r=Selected number of objects

Given,

There are ten number of seats

n=10

We have to arrange 8 students so r=8.

We need to arrange eight students in ten seats in a row.

So n=10, r=8

[tex]10P_{8} =\frac{10!}{(10-8)!}[/tex]

10P₈=10!/2!

=10×9×8×7×6×5×4×3×2/2

=10×9×8×7×6×5×4×3

=1814400

Hence in 1814400 ways can 8 students be assigned a seat in a classroom if there are 10 seats in a row.

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Answer:

Two lines are perpendicular then they cut at right angles to each other.

But, when two lines are parallel then they can never meet to each other.

Step-by-step explanation:

We are given that Martin drew a pair of perpendicular lines and a pair of a parallel lines.

We have to find which statement best describe these statements best compares the pairs of perpendicular and parallel lines.

We know that

Perpendicular lines: That lines which  intersect at right angles to each other.

Parallel lines: That lines which can never meet when the lines produced infinitely.

Hence, a pair of perpendicular lines intersect at right angles and a pair of parallel lines never meet to each  other.

Jessie's total bus ride time, represented by y, is 5 minutes longer than [tex]\frac{2}{3}[/tex] of Robert's bus ride time, which is represented by x, leading to the equation y = ([tex]\frac{2}{3}[/tex])x + 5.

Jessie's bus ride to school is 5 minutes longer than [tex]\frac{2}{3}[/tex] the time of Robert's bus ride. Given that Jessie's time riding the bus is represented by y, and Robert's time riding the bus is represented by x, the equation to represent this situation is:

y = ([tex]\frac{2}{3}[/tex])x + 5.

This equation indicates that if you take [tex]\frac{2}{3}[/tex] of Robert's ride time (x) and then add 5 minutes, you'll get Jessie's bus ride time (y).

Answer:

919.5

Step-by-step explanation:

Final answer:

To solve the equation 4kx+10kx=7 for x, combine like terms to get 14kx = 7, then isolate x by dividing both sides by 14k, resulting in x = 1 / (2k).

Explanation:

The question asks: what is 4kx+10kx=7 solve for x. To solve this equation for x, we first need to combine like terms. We combine 4kx and 10kx to get 14kx. The equation then becomes 14kx = 7. To isolate x, we divide both sides of the equation by 14k. This gives us x = 7 / (14k). It simplifies further to x = 1 / (2k). Therefore, the solution to the equation is x = 1 / (2k).

56.25 / 75 = 0.75 for each can

0.75 *3 = $2.25 for 3 cans

To simplify expressions with negative exponents, rewrite the negative exponent as the reciprocal of the base raised to the positive exponent.

When simplifying expressions with negative exponents, you can use the rule that states a negative exponent can be rewritten as the reciprocal of the base raised to the positive exponent. For example, x^-3 can be rewritten as 1/x^3. You can use this rule with negative exponents in the numerator or denominator, as well as with negative exponents inside parentheses. Here’s an example:

8x^-2 / (2y^-3) = 8 / (2y^3x^2)

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