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maria [59]
3 years ago
9

Albert works at Zippy Burger. Today he is assigned to box fries during the lunch hour. His manager randomly weighs 10 of his lar

ge fries and finds that, on average, he is 'over-stuffing' them by 1.5 oz each. It costs the restaurant 5.2¢ per oz for fries. How much will Zippy Burger lose if 484 orders of large fries are made by Albert this lunch hour? (Assume he overstuffs every large fry by 1.5 oz.)
Mathematics
1 answer:
Georgia [21]3 years ago
4 0
For this we can do 5.2 cents times 1.5 to determine how much zippy is costing the company per box extra so .052 x 1.5 is roughly 7.8 cents per box. So now 7.8 cents Times 484, .078 x 484 is roughly $37.752 cents or $37.75
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The diameter of steel rods manufactured on two different extrusion machines is being investigated. Two random samples of sizes
devlian [24]

Answer:

(a) There is no evidence to support the claim that the two machines produce rods with different mean diameters.

P-value is 0.818

(b) 95% confidence interval for the difference in mean rod diameter is (-0.17, 0.27).

This interval shows that the difference in mean is between -0.17 and 0.27.

Step-by-step explanation:

(a) Null hypothesis: The two machines produce rods with the same mean diameter.

Alternate hypothesis: The two machines produce rods with different mean diameter.

Machine 1

mean = 8.73

variance = 0.35

n1 = 15

Machine 2

mean = 8.68

variance = 0.4

n2 = 17

pooled variance = [(15-1)0.35 + (17-1)0.4] ÷ (15+17-2) = 11.3 ÷ 30 = 0.38

Test statistic (t) = (8.73 - 8.68) ÷ sqrt[0.38(1/15 + 1/17)] = 0.05 ÷ 0.218 = 0.23

degree of freedom = n1+n2-2 = 15+17-2 = 30

Significance level = 0.05 = 5%

Critical values corresponding to 30 degrees of freedom and 5% significance level are -2.042 and 2.042.

Conclusion:

Fail to reject the null hypothesis because the test statistic 0.23 falls within the region bounded by the critical values -2.042 and 2.042.

There is no evidence to support the claim that the two machines produce rods with different mean diameters.

Cumulative area of test statistic is 0.5910

The test is a two-tailed test.

P-value = 2(1 - 0.5910) = 2×0.409 = 0.818

(b) Difference in mean = 8.73 - 8.68 = 0.05

pooled sd = sqrt(pooled variance) = sqrt(0.38) = 0.62

Critical value (t) = 2.042

E = t×pooled sd/√n1+n2 = 2.042×0.62/√15+17 = 0.22

Lower limit of difference in mean = 0.05 - 0.22 = -0.17

Upper limit of difference in mean = 0.05 + 0.22 = 0.27

95% confidence interval for the difference in mean rod diameter is between a lower limit of -0.17 and an upper limit of 0.27.

3 0
4 years ago
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11111nata11111 [884]
The answer is 12 and -7.

12 * -7 = 84

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5 0
3 years ago
Evaluate the expression for the given value of the variable(s).
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hjlf

Given :

The length of the rectangle, l = 8 cm/

The width of the rectangle, w = 75% of 8 cm

                                                 $= \frac{75}{100} \times 8$

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                                                 = 6

Therefore, the width of the rectangle is 6 cm.                                          

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