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2 years ago

What inequality is graphed on the number line below?

Mathematics

2 answers:

vesna_86 [32]2 years ago

5 0

ANSWER: The inequality graphed on the number line is: C. b < 3

vazorg [7]2 years ago

4 0

Answer:

C) b ≤ 3

Step-by-step explanation:

The endpoint of the graph is the midpoint between 2 and 4, so it is 3.

This point is included since indicated as full circle.

The graph includes the line to the left of 3:

b ≤ 3 is the correct one

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PLEASE HURRY ON A TEST!! Given similar figures, find the length of the missing side.

liubo4ka [24]

Answer:

It would be A.10

Step-by-step explanation:

20/5 = 4

9/2.25= 4

40/10 = 4

6 0

3 years ago

Read 2 more answers

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

Polly bought a cracker for $7 and then bought a parrot for $2. By how much did Polly's account change after her transactions?

marysya [2.9K]

Answer:

$Changes = -\$9$

Step-by-step explanation:

Given

$Cracker = \$7$

$Parrot = \$2$

Required

Determine the changes in the account

First, we need to determine the total amount spent;

$Total = Cracker + Parrot$

$Total = \$7 + \$2$

$Total = \$9$

Hence, the changes in her account is a debit of $9 i.e. -$9

7 0

3 years ago

Bas_tet [7]

Answer:

a) We reject H₀ We find difference between the mean of the two groups

b) we accept H₀ : There is no statistical difference ( in the scale of pain) when the difference in samples means is 1

Step-by-step explanation:

Group 1: Only with educational material

Sample size n₁ = 42

Sample mean x₁ = 3,2

Sample standard deviation s₁ = 2,3

Group 2: With exercises with the treatment of individuals

Sample size n₂ = 42

Sample mean x₂ = 5,2

Sample standard deviation s₂ = 2,3

Test Hypothesis

Null Hypothesis H₀ x₂ - x₁ = 0 or x₁ = x₂

Alternative Hypothesis Hₐ x₂ - x ₁ > 0 or x₂ > x₁

Significance level α = 0,01

Sample sizes are n₁ = n₂ > 30

We use z-test

for α = 0,01 z(c) ≈ 2,32

To calculate z(s)

z(s) = ( x₂ - x₁ ) / √ (2,3)²/42 + (2,3)²/42

z(s) = ( 5,2 - 3,2 )/ √0,2519

z(s) = 2 / 0,5

z(s) = 4

Comparing z(s) and z(c)

z(s) > z(c)

z(s) is in the rejection region. We reject H₀ the true average pain level for the control condition exceeds that for the treatment condition.

b) Does the true average pain level for the control condition exceed that for the treatment condition by more than 1.

In this case

z( s) = x₂ - x₁ / 0,5

z(s) = 1 /0,5

z(s) = 2

Comparing z(s) and z(c) now

z(s) < z(c) 2 < 2,32

And z(s) is in the acceptance region ( for the same significance level) and we should accept H₀ equivalent to say that the two groups statistical have the same mean

5 0

3 years ago

Y=-4x-7, x+y=2 please help rn

Mashcka [7]

Use substitution
x -4x - 7 = 2
-3x = 9, x = -3
Y = -4(-3) - 7
Y = 12 - 7, y = 5
Solution: x = -3, y = 5

5 0

3 years ago

Read 2 more answers