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3 years ago

Suppose x=0.5(300-x). what is the value of x?

Mathematics

1 answer:

Sonbull [250]3 years ago

6 0

Expand

x = 150 - 0.5x

Add 0.5x to both sides

x + 0.5x = 150

Simplify x + 0.5x to 1.5x

1.5x = 150

Divide both sides by 1.5

x = 150/1.5

Simplify 150/1.5 to 100

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Simplify sin^2y/sec^2 y−1 to a single trigonometric function

liubo4ka [24]

Answer:

$\frac{ { \sin}^{2} y}{ { \sec}^{2}y - 1 } = { \cos }^{2} y$

Step-by-step explanation:

We know that ${ \tan }^{2} y = { \sec }^{2} y - 1$

Also , ${ \tan}^{2} y = \frac{ { \sin }^{2} y}{ { \cos }^{2}y }$

So ,

$\frac{ { \sin }^{2}y }{ { \sec }^{2}y - 1 } = \frac{ { \sin}^{2} y}{ { \tan }^{2} y} = \frac{ { \sin }^{2}y }{ \frac{ { \sin}^{2} y}{ { \cos}^{2}y } } = { \cos }^{2} y$

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3 years ago

Ivan evaluated the expression 8+12÷3×2 and got 10. What mistake did he make?

lesya [120]

Ivan forgot to follow PEMDAS
Division first
12/3 = 4
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Hope this helps!

3 0

3 years ago

Read 2 more answers

Solve for r -4(r+6)= -49

tatiyna

Answer: r= 25/3

Step-by-step explanation:

r-4(r+6)=-49 distribute the 4

r-4r-24=-49 combine like terms

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-3r=-25 divide by -3

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7 0

4 years ago

Hi its nice to meet you

klio [65]

Answer:

19.5

Step-by-step explanation:

165 divided by 8 is 19.5

8 0

3 years ago

Read 2 more answers

The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer ra

Snezhnost [94]

Let $X$ denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by $X_1,\ldots,X_{36}$ , each independently and identically distributed with distribution $X_i\sim\mathcal N(26,7.2)$ .

You want to find

$\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)$

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

$\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)$

Recall that if $X\sim\mathcal N(\mu,\sigma)$ , then the sampling distribution $\overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right)$ with $n$ being the size of the sample.

Transforming to the standard normal distribution, you have

$Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}$

so that in this case,

$Z=6\dfrac{\overline X-26}{7.2}$

and the probability is equivalent to

$\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)$
$=\mathbb P(Z>1.481)\approx0.0693$

5 0

3 years ago