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saul85 [17]
4 years ago
8

A 65​-inch board is cut into two pieces. One piece is four times the length of the other. Find the lengths of the two pieces.

Mathematics
1 answer:
Novosadov [1.4K]4 years ago
7 0
The shorter one is 13 and the other one is 52.
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6 large taxis x 7 people per large taxi = 42 people. In total there are 75 people. 75-42=32 people left. If you want to divide these people over small taxis, you divide 32 by 4, 32/4=8, so you need 8 small taxis
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PLEASEE HELPP MEEEE!!!!!!!!!<br><br> what is the value of x?
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Step-by-step explanation:

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Y= 34 x - 84 as function notation
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Malik’s recipe for 4 servings of a certain dish requires 1/{1}/{2} cups of pasta. According to this recipe, what is the number o
Ostrovityanka [42]

Answer:

The answer to the question is

1\frac{1}{2} cups

Step-by-step explanation:

According to the recipe number of cups of recipe of past required = 1\frac{1}{2} cups

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In the original question, the number of cups of pasta that Malik will use the next time he prepares this dish is  1\frac{1}{2} cups

3 0
3 years ago
A researcher wishes to estimate with 95% confidence, the proportion of the people who own a home computer. A previous study show
vitfil [10]

Answer:

The minimum sample size necessary is 2305.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

For this problem, we have that:

\pi = 0.4

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The researcher wishes to be accurate within 2% of the true proportion. Find the minimum sample size necessary?

We need a sample size of n.

n is found when M = 0.02. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.02 = 1.96\sqrt{\frac{0.4*0.6}{n}}

0.02\sqrt{n} = 1.96*\sqrt{0.4*0.6}

\sqrt{n} = \frac{1.96*\sqrt{0.4*0.6}}{0.02}

(\sqrt{n})^{2} = (\frac{1.96*\sqrt{0.4*0.6}}{0.02})^{2}

n = 2304.96

Rounding up

The minimum sample size necessary is 2305.

8 0
3 years ago
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