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2 years ago

13

A bodybuilder gains 1.5 pounds each week for 5 weeks. Which equation represents the change in the bodybuilder's weight after the

5 weeks?

1 answer:

astra-53 [7]2 years ago

5 0

Answer:

Step-by-step explanation:

The bodybuilder gains 1.5 each week for 5 weeks.

If the bodybuilder gains 1.5 for 5 weeks, how many pounds did he gain?

1.5 * 5 = 7.5

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Geometry math question

Mamont248 [21]

JK = JH + HK

JK = 82

JH = 22

So, HK = 82 - 22 = 60, or choice (C).

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3 years ago

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What is a real life example of trigonometric ratios

Brut [27]

Answer:

Trigonometry can be used to measure the height of a building or mountains

Step-by-step explanation:

if you know the distance from where you observe the building and the angle of elevation you can easily find the height of the building. Similarly, if you have the value of one side and the angle of depression from the top of the building you can find and another side in the triangle, all you need to know is one side and angle of the triangle.

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2 years ago

Morris has 2 3/4 gallons of ice tea. He gives 3/7 of it to her friend. How many gallons of ice tea does Morris have left?

lara31 [8.8K]

Answer:

$2 \frac{9}{28}$ gallons of ice tea is left with Morris

Step-by-step explanation:

Morris has 2 3/4 gallons of ice tea. He gives 3/7 of it to her friend.

Morris has 2 3/4 gallons of ice tea

$2\frac{3}{4}= \frac{2*4+3}{4}=\frac{11}{4}$

He gives 3/7 of it to her friend.

To find how many gallons of ice tea does Morris have left , we subtract 3/7 from 11/4

$\frac{11}{4}- \frac{3}{7}$

Make the denominators same

$\frac{11*7}{4*7}- \frac{3*4}{7*4}$

$\frac{77}{28}- \frac{12}{28}$

$\frac{65}{28}$

$\frac{65}{28} = 2 \frac{9}{28}$

7 0

2 years ago

Write an expression for "7 divided by <br> c."

Harman [31]

An expression for “7 divided by c” is 7/c or 7 ÷ c

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3 years ago

A shoe manufacturer was investigating the weights of men's soccer cleats. He felt that the weight of these cleats was less than

aleksklad [387]

Answer:

The conclusion is that the researcher was correct

Step-by-step explanation:

From the question we are told that

The sample size is $n = 13$

The sample mean is $\= x = 9.63$

The standard deviation is $s = 0.585$

The significance level is $\alpha = 0.05$

The Null Hypothesis is $H_o : \mu = 0$

The Alternative Hypothesis is $H_a = \mu < 10$

The test statistic is mathematically represented as

$t = \frac{\= x - \mu }{\frac{s}{\sqrt{n} } }$

Substituting values

$t = \frac{9.63 - 10 }{\frac{0.585}{\sqrt{13} } }$

$t = - 2.280$

Now the critical value for $\alpha$ is

$t_{\alpha } = 1.645$

This obtained from the critical value table

So comparing the critical value of alpha and the test value we see that the test value is less than the critical value so the Null Hypothesis is rejected

The conclusion is that the researcher was correct

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3 years ago