8/5 * 5/9 greater than equal to or less than 5/9

Ill mzrkb brainlist plss help

Anvisha [2.4K]

Answer:

i think 41.3

Step-by-step explanation:

6 0

2 years ago

-5(-6k + 10) =<br> Help ASAP

Fudgin [204]

Answer:

Step-by-step explanation:

-5 (-6k + 10)

= 30k - 50

8 0

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

7 0

3 years ago

What is the volume of a hemisphere with a radius of 4.9 ft, rounded to the nearest

Alexxx [7]

Answer:

Step-by-step explanation:

To round 4.9 to nearest tenth means to round the numbers so you only have one digit in the fractional part. 4.9 already has only one digit in the fractional part. Thus, 4.9 is already rounded as much as possible to the nearest tenth and the answer is: 4.9

3 0

2 years ago

Read 2 more answers

Write an expression that represents the perimeter of the football field. Let x represent the length of the football field. Inclu

dlinn [17]

Expression with parenthesis: = 2 ( l + w)
Expression without parenthesis = 2l + 2w

The property we used is known as "Distributive Property"

Let, width = w
Length = w + 70

Perimeter would be: P = 2(w+70+w)
p = 2(2w+70)
p = 4w + 140

Hope this helps!

6 0

3 years ago

8/5 * 5/9 greater than equal to or less than 5/9​

8/5 * 5/9 greater than equal to or less than 5/9