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Ludmilka [50]
3 years ago
9

How to solve 10 = 1\3( x - 15 )

Mathematics
1 answer:
Alona [7]3 years ago
4 0

Answer:

x=45

Step-by-step explanation:

In order to solve for x, you would need to get the variable by itself. First, we would divide 1/3 on both sides.

10 = 1/3 (x - 15)

10 / (1/3) = (x - 15)

30 = x - 15

Then we would add 15 to both sides.

45 = x

Then, we can check our work by plugging the value of x into the original equation.

10 = 1/3 (x - 15)

10 = 1/3 (30)

10=10

The correct answer is x = 45

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Solve the inequality below. Enter your answer as an inequality, such as x < 2/3
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Divide each term by -5 and simplify

Step-by-step explanation

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Write an equation for the following word problem: "three times the sum of a number and ten is 24"
STatiana [176]

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Step-by-step explanation:

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A researcher asks random samples of residents of two separate counties as to whether they had purchased organically grown food i
Leni [432]

Answer:

2 proportions z test

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Step-by-step explanation:

This is testing hypothesis about the difference between two proportions.

When the proportions are tested if they are the test statistic

z= ( p^1-p^2)- (p1-p2) / √p₁q₁/n₁ + p₂q₂/ n₂

where p^1 is the proportion of success in the first sample and   p^2 of size n₁   is the proportion of success in the second sample of size n₂ with unknown proportions of successes p1 and p2 respectively.

When the sample sizes are sufficiently large

z= ( p^1-p^2)- (p1-p2) / √p₁q₁/n₁ + p₂q₂/ n₂ is approximately standard normal.

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7 0
3 years ago
Historically, a certain region has experienced 92 thunder days annually. (A "thunder day" is day on which at least one instance
DiKsa [7]

Answer:

We conclude that the mean number of thunder days is less than 92.

Step-by-step explanation:

We are given that Historically, a certain region has experienced 92 thunder days annually.

Over the past fifteen years, the mean number of thunder days is 72 with a standard deviation of 38.

<u><em>Let </em></u>\mu<u><em> = mean number of thunder days.</em></u>

So, Null Hypothesis, H_0 : \mu \geq 92 days     {means that the mean number of thunder days is more than or equal to 92}

Alternate Hypothesis, H_A : \mu < 92 days     {means that the mean number of thunder days is less than 92}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

                       T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean number of thunder days = 72

             s = sample standard deviation = 38

            n = sample of years = 15

So, <u><em>test statistics</em></u>  =  \frac{72-92}{\frac{38}{\sqrt{15} } }  ~ t_1_4

                               =  -2.038

The value of z test statistics is -2.038.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the t table gives critical value of -1.761 at 14 degree of freedom for left-tailed test.

Since our test statistics is less than the critical value of t as -2.038 < -1.761, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that the mean number of thunder days is less than 92.

7 0
3 years ago
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