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3 years ago

Martin is 6 years younger than his sister. The sum of their ages is no more than 22 years.

Mathematics

1 answer:

Kobotan [32]3 years ago

5 0

So their age can be no more than 22. And Martin is 6 years younger, which means you subtract.

22 ≥ x + x - 6

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Multiplied by 3/4 and then divided by 3/5 is the same as dividing by what?

netineya [11]

We have: x × $\frac{3}{4}$ ÷ $\frac{3}{5}$ = x × $\frac{3}{4}$ × $\frac{5}{3}$ = x × $\frac{5}{4}$ = x ÷ $\frac{4}{5}$

ANSWER: E. 4/5

Ok done. Thank to me :>

6 0

2 years ago

3n-5=-48-40n. Please help . I need the steps to solve and the answer

liraira [26]

3n-5=-48-40n

Move -40n to the other side. Sign changes from -40n to +40n.

3n+40n-5=-48= -48-40n+40n

3n+40n-5=-48

Move -5 to the other side.

3n+40n-5+5=-48+5

3n+40n=-43

43n=-43

Divide by 43 for both sides

43n/43=-43/43

n=-1

Answer: n=-1

5 0

3 years ago

Read 2 more answers

Divide 28 cans of soda into two groups so the ratio is 3 to 4 to

Brums [2.3K]

28 = 3x+4x
x = 4

3x = 12
4x = 16
12:16 = 3:4

6 0

3 years ago

The mean number of words per minute (WPM) read by sixth graders is 8888 with a standard deviation of 1414 WPM. If 137137 sixth g

Bingel [31]

Noticing that there is a pattern of repetition in the question (the numbers are repeated twice), we are assuming that the mean number of words per minute is 88, the standard deviation is of 14 WPM, as well as the number of sixth graders is 137, and that there is a need to estimate the probability that the sample mean would be greater than 89.87.

Answer:

"The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

Step-by-step explanation:

This is a problem of the distribution of sample means. Roughly speaking, we have the probability distribution of samples obtained from the same population. Each sample mean is an estimation of the population mean, and we know that this distribution behaves normally for samples sizes equal or greater than 30 $\\ n \geq 30$ . Mathematically

$\\ \overline{X} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$ [1]

In words, the latter distribution has a mean that equals the population mean, and a standard deviation that also equals the population standard deviation divided by the square root of the sample size.

Moreover, we know that the variable Z follows a normal standard distribution, i.e., a normal distribution that has a population mean $\\ \mu = 0$ and a population standard deviation $\\ \sigma = 1$ .

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [2]

From the question, we know that

The population mean is $\\ \mu = 88$ WPM
The population standard deviation is $\\ \sigma = 14$ WPM

We also know the size of the sample for this case: $\\ n = 137$ sixth graders.

We need to estimate the probability that a sample mean being greater than $\\ \overline{X} = 89.87$ WPM in the distribution of sample means. We can use the formula [2] to find this question.

The probability that the sample mean would be greater than 89.87 WPM

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{89.87 - 88}{\frac{14}{\sqrt{137}}}$

$\\ Z = \frac{1.87}{\frac{14}{\sqrt{137}}}$

$\\ Z = 1.5634 \approx 1.56$

This is a standardized value and it tells us that the sample with mean 89.87 is 1.56 standard deviations above the mean of the sampling distribution.

We can consult the probability of P(z<1.56) in any cumulative standard normal table available in Statistics books or on the Internet. Of course, this probability is the same that $\\ P(\overline{X} < 89.87)$ . Then