300 reduced by 3 times x is 33

Kat owns 15% more figurines than she did last year. She purchased F figurines this year. Which of the following expressions coul

Stella [2.4K]

Answer: F / 0.15 = 100F / 15 = 20F / 3 ≈ 6.67F.

Explanation:

1) The three expresssions, F / 0.15; 15 F / 100; 3F / 20; are equivalent.

2) The number of figurines Kat owned last year may be expressed in terms of the number the of figurines purchased, F, this year, and the 15%, following this procedure:

Call n the amount of figurines owned last year.

F is the number of figurines purchased: F = 15%n = 0,15n

Hence, you just need to clear F from the expression F = 0.15n

Divide both sides by 0.15:

F / 0.15 = n or n = F / 0.15 (reflexive property)

To prove the equialavence of the other expresssions you just have to use the properties of multiplication or division:

(F / 0.15) × (100 / 100) = 100F / 15

100F / 15 = (100/5)F / (15/5) = 20F / 3 ≈ 6.67F

4 0

3 years ago

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If you had to put up a fence around a circular baseball field, you would need to know the _________?

laila [671]

You would need to know the diameter of the field, and from that, you need to get the circumference of the field by using the formula C = πd.

By getting the circumference, you can therefore determine the perimeter/length of the fence you will need to enclose it.

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3 years ago

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PLEASE HELP ASAP!! WILL MARK BRAINLIEST!!

kipiarov [429]

It’s B. Dan is 124 miles from home at noon and is traveling at 62 miles per hour

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3 years ago

You work in the HR department at a large franchise. you want to test whether you have set your employee monthly allowances corre

ra1l [238]

Answer:

1) Null hypothesis: $\mu \leq 500$

Alternative hypothesis: $\mu > 500$

$z=\frac{640-500}{\frac{150}{\sqrt{40}}}=5.90$

For this case since we are conducting a right tailed test we need to find a critical value in the normal standard distribution who accumulates 0.01 of the area in the right and we got:

$z_{crit}= 2.33$

For this case we see that the calculated value is higher than the critical value

Since the calculated value is higher than the critical value we have enugh evidence to reject the null hypothesis at 1% of significance level

2) Since is a right tailed test the p value would be:

$p_v =P(z>5.90)=1.82x10^{-9}$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, same conclusion for part 1

Step-by-step explanation:

Part 1

Data given

$\bar X=640$ represent the sample mean

$\sigma=150$ represent the population standard deviation

$n=40$ sample size

$\mu_o =500$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

Step1:State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is higher than 500, the system of hypothesis would be:

Null hypothesis: $\mu \leq 500$

Alternative hypothesis: $\mu > 500$

Step 2: Calculate the statistic

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

We can replace in formula (1) the info given like this:

$z=\frac{640-500}{\frac{150}{\sqrt{40}}}=5.90$

Step 3: Calculate the critical value

For this case since we are conducting a right tailed test we need to find a critical value in the normal standard distribution who accumulates 0.01 of the area in the right and we got:

$z_{crit}= 2.33$

Step 4: Compare the statistic with the critical value

For this case we see that the calculated value is higher than the critical value

Step 5: Decision

Since the calculated value is higher than the critical value we have enugh evidence to reject the null hypothesis at 1% of significance level

Part 2

P-value

Since is a right tailed test the p value would be:

$p_v =P(z>5.90)=1.82x10^{-9}$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, same conclusion for part 1

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The relationship between the number of dollars per month that a company spends on

Alex17521 [72]

C)$10000. Hola chao

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3 years ago

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