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3 years ago

Helppp can someone solve this please

Mathematics

1 answer:

Norma-Jean [14]3 years ago

5 0

Answer:

$\displaystyle x=-15$

Step-by-step explanation:

Solution Of A System Of Equations

A system of linear equations is given as

$\displaystyle \left\{\begin{matrix}ax+by=c\\ dx+ey=f\end{matrix}\right.$

There are many methods to solve them. We will use the method of reduction

The given system is

$\displaystyle \left\{\begin{matrix}2x+3y=45\\ x+y=10\end{matrix}\right.$

Multiplying the second equation by -3

$\displaystyle \left\{\begin{matrix}2x+3y=45\\ -3x-3y=-30\end{matrix}\right.$

Adding the resulting equations

$\displaystyle -x=15$

$\displaystyle x=-15$

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A department store, on average, has daily sales of 28,651.79. the standard deviation pf sales is $1000. On Tuesday, the store so

adelina 88 [10]

Answer:

Tuesday's Z-score is 7.56

Step-by-step explanation:

We are given that a department store, on average, has daily sales of 28,651.79 and the standard deviation of sales is $1000.

Also, it is given that on Tuesday, the store sold $36,211.08 worth of goods.

Let X = Daily sales of goods

So, X ~ N( $\mu = 28,651.79, \sigma^{2} =1000^{2}$ )

The z-score probability distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ standard normal N(0,1)

Now, Tuesday,s Z-score is given by;

Z = $\frac{36,211.08-28,651.79}{1000}$ = 7.56

Yes, Tuesday was an unusually good day as on this day more worth of sales takes place as compared to the average daily sales of $28,651.79 .

5 0

4 years ago

UCF believes that the average time someone spends in the gym is 56 minutes. The university statistician takes a random sample of

11111nata11111 [884]

Answer:

We conclude that the average time someone spends in the gym is different from 56 minutes.

Step-by-step explanation:

We are given that UCF believes that the average time someone spends in the gym is 56 minutes.

The university statistician takes a random sample of 32 gym goers and finds the average time of the sample was 50 minutes. Assume it is known the standard deviation of time all people spend in the gym is 8 minutes.

Let $\mu$ = population average time someone spends in the gym

SO, Null Hypothesis, $H_0$ : $\mu$ = 56 minutes {means that the average time someone spends in the gym is 56 minutes}

Alternate Hypothesis, $H_A$ : $\mu\neq$ 56 minutes {means that the average time someone spends in the gym is different from 56 minutes}

The test statistics that will be used here is One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average time someone takes in the gym = 50 min

$\sigma$ = population standard deviation = 8 minutes

n = sample of gym goers = 32

So, test statistics = $\frac{50-56}{\frac{8}{\sqrt{32} } }$

= -4.243

Since in the question we are not given the level of significance so we assume it to b 5%. Now at 5% significance level, the z table gives critical value between -1.96 and 1.96 for two-tailed test. Since our test statistics does not lie within the range of critical values of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the average time someone spends in the gym is different from 56 minutes.

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Answer:

Step-by-step explanation:

graph_1 and graph_2 are for this task

graph_3 is for previous task.