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3 years ago

Find the value of x that makes each equation true.

1 answer:

8 0

$1.\\\\ 3x-10=2\\3x=2+10\\3x=12\ \ /:3\\ x=\frac{12}{3}\\x=4\\\\2.\\\\ 8x-10=x+11\\8x-x=11+10\\7x=21\ \ /:7\\x=\frac{21}{7}\\x=7$

$3.\\\\ 5x-7=7x-17\\5x-7x=-17+7\\-2x=-10\ \ /:(-2)\\x= \frac{-10}{-2}\\x=5$

$4.\\\\ 3x+9=44-2x\\3x+2x=44-9\\5x=35\ \ /:5\\x=\frac{35}{7}\\x=5\\\\5. \\\\3x+6=x+2\\3x-x=2-6\\2x=-4\ \ /:2\\x=-\frac{4}{2}\\x=-2
$

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Help me ugent <br> 12 x 12 ^ 1

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Answer:

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Step-by-step explanation:

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2 years ago

Every US. Citizen has a 9 digit social security number. The first 3 digits are a geographic code, the next two digits are determ

Ilia_Sergeevich [38]

Probability of success =

   (number of possibilities that would be considered successful)
divided by
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Total number of different possible outcomes =

   total number of different possible values of the last 2 digits

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   = 100 of them .

The number of possibilities that will match your age = only 1 .

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2 years ago

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almond37 [142]

Answer:

147?

Step-by-step explanation:

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2 years ago

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

2 years ago

Let f(x)=3x2-2 wrote a function g whose graph is a reflection in the y axis of the graph of f

lorasvet [3.4K]

9514 1404 393

Answer:

g(x) = 3x^2 -2

Step-by-step explanation:

Reflection in the y-axis replaces x with -x.

g(x) = f(-x) = 3(-x)^2 -2

g(x) = 3x^2 -2

_____

The given function is symmetrical about the y-axis, so its reflection is itself.

6 0

2 years ago