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nignag [31]
3 years ago
11

Lenox ironed of the shirts over the weekend. She plans to split the remainder of the work equally over the next 5 evenings. If L

enox has 40 shirts, how many shirts will she need to iron on Thursday and Friday? Do not include units in your answer.
Mathematics
1 answer:
azamat3 years ago
3 0

Answer: 16

Step-by-step explanation:

Total shirts = 40

Number of evenings = 5

If we split the remainder of the work equally over the next 5 evenings.

then, the number of shirts she will need to iron on each evening = \dfrac{40}{5}=8

Number of shirts need to buy on Thursday and Friday = 2 (8) = 16

Hence, the number of shirts she need to iron on Thursday and Friday = 16

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Answer:

The frequency table is listed down below.

Step-by-step explanation:

1:1

2:1

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9:1

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7 0
4 years ago
Work out the precentage change when a price of 40 is decreased to 38
aksik [14]

Answer:

5%

Step-by-step explanation:

38 is 95% of 40 therefore the change is 5%

8 0
3 years ago
Which of the following expressions are equivalent to - ( -5/3)
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3 0
3 years ago
Read 2 more answers
Mr. Good Wrench advertises that a customer will have to wait no more than 30 minutes for an oil change. A sample of 26 oil chang
Andru [333]

Answer:

The 90% confidence interval for the population standard deviation waiting time for an oil change is (3.9, 6.3).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the population standard deviation is:

CI=\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2, (n-1)}}}\leq \sigma\leq \sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2, (n-1)}}}

The information provided is:

<em>n</em> = 26

<em>s</em> = 4.8 minutes

Confidence level = 90%

Compute the critical values of Chi-square as follows:

\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.10/2, (26-1)}=\chi^{2}_{0.05, 25}=37.652

\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.10/2, (26-1)}=\chi^{2}_{0.95, 25}=14.611

*Use a Chi-square table.

Compute the 90% confidence interval for the population standard deviation waiting time for an oil change as follows:

CI=\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2, (n-1)}}}\leq \sigma\leq \sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2, (n-1)}}}

     =\sqrt{\frac{(26-1)\times 4.8^{2}}{37.652}}\leq \sigma\leq \sqrt{\frac{(26-1)\times 4.8^{2}}{14.611}}\\\\=3.9113\leq \sigma\leq 6.2787\\\\\approx 3.9 \leq \sigma\leq6.3

Thus, the 90% confidence interval for the population standard deviation waiting time for an oil change is (3.9, 6.3).

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Rapide s’il vous plaît
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