Can someone give me a linear equation word problem, and the answer to it

1 answer:

5 0

Raymond just got done jumping at Super Bounce Trampoline Center. The total cost of his session was \$43.25$43.25dollar sign, 43, point, 25. He had to pay a \$7$7dollar sign, 7 entrance fee and \$1.25$1.25dollar sign, 1, point, 25for every minute he was on the trampoline.Write an equation to determine the number of minutes (t)(t)left parenthesis, t, right parenthesis that Raymond was on the trampoline.

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When I sit and watch my students take exams, I often think to myself "I wonder if students with bright calculators are impacted

hodyreva [135]

Answer:

There is a difference between the two means.

Step-by-step explanation:

The hypothesis can be defined as:

H₀: The mean exam scores of my SAT 215 students with colorful calculators are same as the mean scores of my STA 215 students with plain black calculators, i.e. μ₁ - μ₂ = 0.

Hₐ: The mean exam scores of my SAT 215 students with colorful calculators are different than the mean scores of my STA 215 students with plain black calculators, i.e. μ₁ - μ₂ ≠ 0.

Assume that the significance level of the test is, α = 0.05. Also assuming that the population variances are equal.

The decision rule:

A 95% confidence interval for mean difference can be used to determine the result of the hypothesis test. If the 95% confidence interval contains the null hypothesis value, i.e. 0 then the null hypothesis will not be rejected.

The 95% confidence interval for mean difference is:

$CI=\bar x_{1}-\bar x_{2}\pm t_{\alpha/2, (n_{1}+n_{2}-2)}\times S_{p}\times \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}$

Compute the pooled standard deviation as follows:

$S_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}} {n_{1}+n_{2}-2}}}=\sqrt{\frac{(49-1)(4.7)^{2}+(38-1)(5.7)^{2}}{49+38-2}}=5.16$

The critical value of t is:

$t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.05/2, (49+38-2)}=t_{0.025, 85}=1.984$

*Use a t-table.

Compute the 95% confidence interval for mean difference as follows:

$CI=\bar x_{1}-\bar x_{2}\pm t_{\alpha/2, (n_{1}+n_{2}-2)}\times S_{p}\times \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}$

$=(84-87)\pm 1.984\times 5.16\times \sqrt{\frac{1}{49}+\frac{1}{38}}$

$=-3\pm 2.133\\=(-5.133, -0.867)\\\approx(-5.13, -0.87)$

The 95% confidence interval for mean difference is (-5.13, -0.87).

The confidence interval does not contains the value 0. This implies that the null hypothesis will be rejected at 5% level of significance.

Hence, concluding that the mean exam scores of my STA 215 students with colorful calculators are different than the mean scores of my STA 215 students with plain black calculators.

7 0

3 years ago

Division in colone 42735 divided by 5 42735:5

8090 [49]

8547 itu jawabannya dek

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How do I Graph 2y=-8

qwelly [4]

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