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yKpoI14uk [10]
2 years ago
11

Scarlet and Paula are on a vacation. They went to a store to buy some souvenirs for their friends back home. Scarlet bought thre

e pairs of sunglasses and two shirts for $81 and Paula bought one pair of sunglasses and five shirts for $105. What is the cost of one pair of sunglasses and one shirt? The cost of one pair of sunglasses is $15 and the cost of one shirt is $18. The cost of one pair of sunglasses is $18 and the cost of one shirt is $15. The cost of one pair of sunglasses is $12 and the cost of one shirt is $23. The cost of one pair of sunglasses is $23 and the cost of one shirt is $12.
Mathematics
2 answers:
snow_lady [41]2 years ago
8 0

Answer:

a

Step-by-step explanation:

1. Choose a person.

2. Substitute the numbers into the correct values.

105 - 15 = ?/5 = 18

105 - 18 = ?/5 = 3.48

105 - 12 = ?/5 = 18.6

105 - 23 = ?/5 = 16.4

It is A because it is the only one in which the answer is equivalent to the cost.

Hope this helps:)

timama [110]2 years ago
6 0

Answer:

The answer is A. .

Step-by-step explanation:

Let s represent the amount of sunglasses bought and t represent the amount of shirts bought.

Scarlet bought 3 pairs of sunglasses and 2 shirts for $81. In other words:

3s+2t=81

Paula bought one pair of sunglasses and five shirts for $105. In other words:

s+5t=105

We have a system of equations and can solve it by isolating s.

s+5t=105

s=105-5t

Now, substitute.

3s+2t=81

3(105-5t)+2t=81

315-15t+2t=81

-13t=-234

t=18

Now, determine s.

s=105-5t

s=105-5(18)

s=15

Thus, a pair of sunglasses is $15 and a shirt is $18.

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AleksandrR [38]

Answer:

(240-210) -1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=27.953

(240-210) +1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=32.047

And the confidence interval for the difference is between:

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Step-by-step explanation:

We have the following info given:

\bar X_1 = 240 sample mean for medium Pizzas from Prim's

\bar X_2 = 210 sample mean for medium Pizzas from Pizza Place

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s_2 =5.7 sample deviation for Pizza Palca

n_1 =n_2 = 100  sample size selected for each case

The confidence interval for the difference of means is given by:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}

And for the 95% confidence we need a significance level of \alpha=1-0.95=0.05 and \alpha/2 =0.025, the degrees of freedom are given by:

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(240-210) -1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=27.953

(240-210) +1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=32.047

And the confidence interval for the difference is between:

27.953 \leq \mu_1 -\mu_2 \leq 32.047

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