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2 years ago

Jordan made 8th blankets with 136 yard of material how many yards of material does she need to make 23 blanket

Mathematics

2 answers:

prisoha [69]2 years ago

4 0

The answer is three hundred ninety one

miss Akunina [59]2 years ago

3 0

Answer:

391 yds of material

Step-by-step explanation:

to find the answer, you take 136 yards divided by 8 blankets, which equals 17.

Next, you multiply 23 blankets by that 17.

Then you find your answer, 391 yds.

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Skipping Lunch A nutritionist wishes to determine, within 3%, the true proportion of adults who do not eat any lunch. If he wish

mixas84 [53]

Answer:

A sample of at least 545 adults is needed.

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 $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$\pi = 0.15$

The margin of error is:

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

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 $Z = 1.96$ .

If he wishes to be 95% confident that his estimate contains the population proportion, how large a sample will be necessary?

We need a sample of at least n.

n is found when M = 0.03. So

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

$0.03 = 1.96\sqrt{\frac{0.15*0.85}{n}}$

$0.03\sqrt{n} = 1.96\sqrt{0.15*0.85}$

$\sqrt{n} = \frac{1.96\sqrt{0.15*0.85}}{0.03}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.15*0.85}}{0.03})^{2}$

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Rounding up

A sample of at least 545 adults is needed.

6 0

2 years ago

a poll reported that 65% of adults were satisfied with the job the major airlines were doing. Suppose 20 adults are selected at

Triss [41]

Answer:

the probability that 11 adults are satisfied is 0.115 (11.5%)

Step-by-step explanation:

since the result of each adult is independent of others , then since the adults are selected at random , the random variable X= number of adults who are satisfied , has a binomial probability distribution , where

P(X=x) = n!/[(n-x)!*x!] * p^x * (1-p)^(n-x)

where

n= sample size = 20

p= probability of any adult to be satisfied with the job of airlines = 0.65

x = number of adults who are satisfied

P(X=x) = probability that x adults are satisfied

then for x= 11 , we have

P(X=11) = 20!/[(20-11)!*11!] * 0.65^11 * (1-0.65)^(20-11) = 0.115 (11.5%)

then the probability that 11 adults are satisfied is 0.115 (11.5%)

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3 years ago

What percent of 20 is 11

Jet001 [13]

'is' in mathematics means equal
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$\frac{x}{100} \times 20 = 11\\\\\frac{x}{5} = 11\\\\\boxed{\bf{x = 55}}$

55% of 20 is 11.

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3 years ago

Jordan made 8th blankets with 136 yard of material how many yards of material does she need to make 23 blanket​

Jordan made 8th blankets with 136 yard of material how many yards of material does she need to make 23 blanket