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3 years ago

12

Aubrey worked at the concession stand during a recent sporting event at her high school. During the first hour of the event, she

sold 11 orders of nachos and 18 tacos for a total of 64.95 . During the second hour, she sold 6 orders of nachos and tacos 15 for a total of 43.20.

1 answer:

ella [17]3 years ago

8 0

Answer:

cost of nachos = $3.45

Cost of tacos = $1.5

Step-by-step explanation:

Let

x = cost of nachos

y = cost of tacos

11x + 18y = 64.95 (1)

6x + 15y = 43.20 (2)

Multiply (1) by 6 and (2) by 11 to eliminate x

66x + 108y = 389.70

66x + 165y = 475.20

Subtract to eliminate x

165y - 108y = 475.20 - 389.70

57y = 85.50

y = 85.50/57

y = 1.5

Substitute y = 1.5 to find x

11x + 18y = 64.95 (1)

11x + 18(1.5) = 64.95

11x + 27 = 64.95

11x = 64.95 - 27

11x = 37.95

x = 37.95/11

x = 3.45

cost of nachos = $3.45

Cost of tacos = $1.5

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A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past

Vadim26 [7]

Answer:

We conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

Step-by-step explanation:

We are given that a particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8000.

From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9800 miles.

<u><em>Let </em></u> $\mu$ <u><em> = average miles for deluxe tires</em></u>

So, Null Hypothesis, $H_0$ : $\mu \geq$ 50,000 miles {means that deluxe tire averages at least 50,000 miles before it needs to be replaced}

Alternate Hypothesis, $H_A$ : $\mu$ < 50,000 miles {means that deluxe tire averages less than 50,000 miles before it needs to be replaced}

The test statistics that will be used here is <u>One-sample z test statistics</u> as we know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean lifespan = 46,500 miles

$\sigma$ = population standard deviation = 8000 miles

n = sample of tires = 28

So, <u><em>test statistics</em></u> = $\frac{46,500-50,000}{\frac{8000}{\sqrt{28} } }$

= -2.315

The value of the test statistics is -2.315.

Now at 5% significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is less than the critical value of z as -2.315 < -1.6449, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

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Ne4ueva [31]

Answer:

Step-by-step explanation:

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Use the 2nd button and the cos button to find the missing angle.

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3 years ago

$30.50 - $17.79= what<br> Solve this problem in two diffrent ways (5th grade math)

puteri [66]

$30.50-$17.79 is $12.71. I hope that helps you :)

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3 years ago

Read 2 more answers

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Kruka [31]

Answer:

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Step-by-step explanation:

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Hope this helps

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2 years ago

Consider the following hypothesis test. : : The following results are for two independent samples taken from two populations. Ex

I am Lyosha [343]

Answer:

$z = -1.53$ --- test statistic

$p = 0.1260$ --- p value

Conclusion: Fail to reject the null hypothesis.

Step-by-step explanation:

Given

$n_1 = 80$ $\bar x_1= 104$ $\sigma_1 = 8.4$

$n_2 = 70$ $\bar x_2 = 106$ $\sigma_2 = 7.6$

$H_o: \mu_1 - \mu_2 = 0$ --- Null hypothesis

$H_a: \mu_1 - \mu_2 \ne 0$ ---- Alternate hypothesis

$\alpha = 0.05$

Solving (a): The test statistic

This is calculated as:

$z = \frac{\bar x_1 - \bar x_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} }}$

So, we have:

$z = \frac{104 - 106}{\sqrt{\frac{8.4^2}{80} + \frac{7.6^2}{70} }}$

$z = \frac{104 - 106}{\sqrt{\frac{70.56}{80} + \frac{57.76}{70}}}$

$z = \frac{-2}{\sqrt{0.8820 + 0.8251}}$

$z = \frac{-2}{\sqrt{1.7071}}$

$z = \frac{-2}{1.3066}$

$z = -1.53$

Solving (b): The p value

This is calculated as:

$p = 2 * P(Z < z)$

So, we have:

$p = 2 * P(Z < -1.53)$

Look up the z probability in the z score table. So, the expression becomes

$p = 2 * 0.0630$

$p = 0.1260$

Solving (c): With $\alpha = 0.05$ , what is the conclusion based on the p value

We have:

$\alpha = 0.05$

In (b), we have:

$p = 0.1260$

By comparison:

$p > \alpha$

i.e.

$0.1260 > 0.05$

So, we fail to reject the null hypothesis.

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2 years ago