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frez [133]
2 years ago
15

Algebra There were 4 times the number of students in fourth grade at the basketball game.How many students attended the basketba

ll game? Write and solve an equation.

Mathematics
1 answer:
denis-greek [22]2 years ago
8 0

Answer:

200 students attended the basketball game

Step-by-step explanation:

The complete question in the attached figure

Let

x ------> the number of students in fourth grade

s -----> the number of students at the basketball game

we know that

The number of students at the basketball game is four times the number of students in fourth grade

so

The linear expression is

s=4x -----> equation A

x=50\ students\ in\ fourth\ grade -----> equation B

substitute equation B in equation A and solve for y

s=4(50)

s=200\ students\ at\ the\ basketball\ game

therefore

200 students attended the basketball game

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Answer:

We failed to reject H₀

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We failed to reject H₀

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We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

Step-by-step explanation:

Set up hypotheses:

Null hypotheses = H₀: p = 0.40

Alternate hypotheses = H₁: p < 0.40

Determine the level of significance and Z-score:

Given level of significance = 1% = 0.01

Since it is a lower tailed test,

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Determine type of test:

Since the alternate hypothesis states that less than 40% of U.S. cell phone owners use their phone for most of their online browsing, therefore we will use a lower tailed test.

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Since the sample size is quite large (n > 30) therefore, we will use Z-distribution.

Set up decision rule:

Since it is a lower tailed test, using a Z statistic at a significance level of 1%

We Reject H₀ if Z < -1.645

We Reject H₀ if p ≤ α

Compute the test statistic:

$ Z =  \frac{\hat{p} - p}{ \sqrt{\frac{p(1-p)}{n} }}  $

$ Z =  \frac{0.31 - 0.40}{ \sqrt{\frac{0.40(1-0.40)}{100} }}  $

$ Z =  \frac{- 0.09}{ 0.048989 }  $

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Conclusion:

We failed to reject H₀

Z > -1.645

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We failed to reject H₀

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0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

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Answer:

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