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2 years ago

What fraction is bigger 13/15 or 21/25 Really need answer urgently

1 answer:

7 0

$\frac{13}{15}=\frac{65}{75}\\\\\frac{21}{25}=\frac{63}{75}\\\\\frac{65}{75}\ \textgreater \ \frac{63}{75}$

$\frac{13}{15}$ is bigger than $\frac{21}{25}$ .

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HLEP ME PLEASE !!!!!!

Advocard [28]

Answer:

f(x) + g(x) = 3x + 7

Step-by-step explanation:

f(x) = 2x + 2, g(x) = x + 5

f(x) + g(x) = 2x + 2 +x + 5

f(x) + g(x) = 2x +x + 5 +2

f(x) + g(x) = 3x + 7

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3 years ago

As part of a class project, a university student surveyed the students in the cafeteria lunch line to look for a relationship be

Vedmedyk [2.9K]

Answer:

0.63

Step-by-step explanation:

To get the relative frequency of students with red hair, we need to find the relative frequency of students with red hair from the students with gray eyes.

Since we already know that the total number of students with gray eyes is 35 and the number of red hair students with gray eye is 22

The formula we will use to calculate the relative frequency will be

The relative frequency of students with red hair = Total students of red hair with gray eyes divide by total number of gray eye students

22 ÷ 35 = 0.628 ≅ 0.63

6 1

3 years ago

Based on the graph what are the solutions to ax^2 + bx+c=0 Select all that apply

AlexFokin [52]

Answer:

x=-2 and x = 5

Step-by-step explanation:

The solutions to the equation are where the graph crosses the x axis

We can see that the graph crosses at x=-2 and x = 5

8 0

3 years ago

Read 2 more answers

Please help me! Screenshot of question is attached. 15 Points and Brainliest to the correct answer.

ehidna [41]

Hello,

I would love to answer but there is no screenshot to see!

7 0

3 years ago

A sample of 200 observations from the first population indicated that x1 is 170. A sample of 150 observations from the second po

igor_vitrenko [27]

Answer:

a) For this case the value of the significanceis $\alpha=0.05$ and $\alpha/2 =0.025$ , we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:

$z_{\alpha/2} =1.96$

If the calculated statistic $|z_{calc}| >1.96$ we can reject the null hypothesis at 5% of significance

b) Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{170+110}{200+150}=0.8$

c) $z=\frac{0.85-0.733}{\sqrt{0.8(1-0.8)(\frac{1}{200}+\frac{1}{150})}}=2.708$

d) Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.

Step-by-step explanation:

Data given and notation

$X_{1}=170$ represent the number of people with the characteristic 1

$X_{2}=110$ represent the number of people with the characteristic 2

$n_{1}=200$ sample 1 selected

$n_{2}=150$ sample 2 selected

$p_{1}=\frac{170}{200}=0.85$ represent the proportion estimated for the sample 1

$p_{2}=\frac{110}{150}=0.733$ represent the proportion estimated for the sample 2

$\hat p$ represent the pooled estimate of p

z would represent the statistic (variable of interest)

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

$\alpha=0.05$ significance level given

Concepts and formulas to use

We need to conduct a hypothesis in order to check if is there is a difference between the two proportions, the system of hypothesis would be:

Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

a.State the decision rule.

For this case the value of the significanceis $\alpha=0.05$ and $\alpha/2 =0.025$ , we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:

$z_{\alpha/2} =1.96$

If the calculated statistic $|z_{calc}| >1.96$ we can reject the null hypothesis at 5% of significance

b. Compute the pooled proportion.

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{170+110}{200+150}=0.8$

c. Compute the value of the test statistic.

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.85-0.733}{\sqrt{0.8(1-0.8)(\frac{1}{200}+\frac{1}{150})}}=2.708$

d. What is your decision regarding the null hypothesis?

Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.

5 0

2 years ago