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2 years ago

A piece of cloth is 12 yards in length. if 3/4 of the cloth is used for bedsheet, how many yards of the cloth will be left?

Mathematics

1 answer:

larisa [96]2 years ago

3 0

Answer:

9 yards

Step-by-step explanation:

multiply the length of the cloth by the length used for bedsheet

$12*\frac{3}{4}\\9 yards$

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What is 22 percent of 50

Lady_Fox [76]

Translate
'what' means unknown or find this or ?
'is' means = or equals
22%
'of' means multiply
50

percent means parts out of 100
x%=x/100 so 22%=22/100

so

what is 22% of 50 means
?=22/100 times 50
?=1100/100
?=11
answer is 11

the hack, is to recognize that 50 is 1/2 of 100, so 22% of 1/2 of 100 is 22/2=11

answe ris 11

5 0

3 years ago

Read 2 more answers

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of

iris [78.8K]

Answer:

The probability that the mean battery life would be greater than 533.2 minutes (in a sample of 75 batteries) is $\\ P(z>0.48) = P(x>533.2) = 0.3156$

Step-by-step explanation:

The main thing we have to take into account in this question is that we are about to find the probability of a sample mean. The distribution for sample means follows a normal distribution with mean $\\ \mu$ and standard deviation $\\ \frac{\sigma}{\sqrt{n}}$ . Mathematically

$\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

For values of the sample $\\ n \ge 30$ , no matter the distribution the data come from.

And the variable z follows a standard normal distribution, and, as we can remember, this distribution has a mean = 0 and a standard deviation = 1. Mathematically

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [1]

That is

$\\ z \sim N(0, 1)$

We have a variance of 3364. That is, a standard deviation of

$\\ \sigma^2 = 3364; \sigma = \sqrt{3364} = 58$

The population mean is

$\\ \mu = 530$

The sample size is $\\ n = 75$

The sample mean is $\\ \overline{x} = 533.2$

With all this information, we can solve the question

The probability that the mean battery life would be greater than 533.2 minutes

Using equation [1]

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ z = \frac{533.2 - 530}{\frac{58}{\sqrt{75}}}$

$\\ z = \frac{3.2}{\frac{58}{8.66025}}$

$\\ z = \frac{3.2}{6.69726}$

$\\ z = 0.47780$

With this value of z we can consult a cumulative standard normal table (or use some statistic program) to find the cumulative probability for z (and remember that this variable follows a standard normal distribution).

Most standard normal tables have values for z for only two decimals, so we can round the previous value for z as z = 0.48.

Then

$\\ P(z$

However, in the question we are asked for $\\ P(z>0.48) = P(x>533.2)$ . As well as all normal distributions, the standard normal distribution is symmetrical around the mean, and we have

$\\ P(z>0.48) = 1 - P(z$

Thus

$\\ P(z>0.48) = 1 - 0.68439$

$\\ P(z>0.48) = 0.31561$

Rounding to four decimal places, we have

$\\ P(z>0.48) = 0.3156$

So, the probability that the mean battery life would be greater than 533.2 minutes is (in a sample of 75 batteries) $\\ P(z>0.48) = P(x>533.2) = 0.3156$ .

5 0

3 years ago

Leo is saving money to buy a snowboard. In March, he saved

Airida [17]

270 I think but I’m not 100% sure

7 0

3 years ago

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A florist prepares an order of daisies and roses. For every 5 daisies she orders 2 roes. What 2 equations show the relationship

Vsevolod [243]

Answer:

$d = 2.5r$ and $r = \frac{d}{2.5}$