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3 years ago

7

The ratio of teachers to students at a school is 3:74 there are 2500 students at the school about hiw many teachers are at the s

chool

1 answer:

zhannawk [14.2K]3 years ago

8 0

Answer:

The answer is 101. There are 101 teachers.

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Can someone help me, I really need help .

Y_Kistochka [10]

Yes I can help you with the question you have displayed

3 0

2 years ago

Read 2 more answers

Expand the following:<br> a) x(x + 2)<br> b) x(2x - 5)<br> c) 2x(3x + 4)<br> d) 6x(x - 2)

TEA [102]

Answer:

<h2> $a. \: {x}^{2} + 2$ </h2>

<h2> </h2><h2> $b. \: 2 {x}^{2} - 5x$ </h2><h2> </h2><h2> $c. \: 6 {x}^{2} + 8x$ </h2><h3> </h3><h3> $d. \: 6 {x}^{2} - 12x$ </h3>

solution,

$a. \: x(x + 2) \\ \: \: = x \times x + 2 \times x \\ \: \: = {x}^{2} + 2x$

$b . \: x(2x - 5) \\ \: = x \times 2x - x \times 5 \\ \: \: = 2 {x}^{2} - 5x$

$c. \: 2x(3x + 4) \\ \: = 2x \times 3x + 2x \times 4 \\ \: \: = 6 {x}^{2} + 8x$

$d. \: 6x(x - 2) \\ \: \: = 6x \times x - 6x \times 2 \\ \: \: = 6 {x}^{2} - 12x$

<h2> </h2><h2 />

Hope this helps...

Good luck on your assignment..

4 0

2 years ago

127.5 is 250% of what?

Inga [223]

Your answer is 127.5 is 250% of 51

4 0

3 years ago

What is the value of x in the equation x – y = 30, when y = 15?

slamgirl [31]

Since <span><span>−y</span></span> does not contain the variable to solve for, move it to the right side of the equation by adding <span>y</span> to both sides.<span>x=<span>y+<span>30</span></span></span>

5 0

3 years ago

Suppose the weights of the Boxers at this club are Normally distributed with a mean of 166 pounds and a standard deviation of 5.

lesya [120]

Answer:

0.1994 is the required probability.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 166 pounds

Standard Deviation, σ = 5.3 pounds

Sample size, n = 20

We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling =

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{5.3}{\sqrt{20}} = 1.1851$

P(sample of 20 boxers is more than 167 pounds)

$P( x > 167) = P( z > \displaystyle\frac{167 - 166}{1.1851}) = P(z > 0.8438)$

$= 1 - P(z \leq 0.8438)$

Calculation the value from standard normal z table, we have,

$P(x > 167) = 1 - 0.8006= 0.1994 = 19.94\%$

0.1994 is the probability that the mean weight of a random sample of 20 boxers is more than 167 pounds

3 0

2 years ago