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3 years ago

7k -2(3k + 1) - 9 Y’all I need help.

Mathematics

2 answers:

rusak2 [61]3 years ago

6 0

Answer:

k - 11

Step-by-step explanation:

7k - 2(3k +1) - 9

So first we start by multiplying out the brackets. Multiply 2 by 3k and 1 then it becomes:

7k - 6k + 2 - 9

Then collect like terms subtract 6k from 7k and 9 from 2:

k - 11

HOPE THIS HELPED

larisa86 [58]3 years ago

4 0

Answer:

k - 11

Step-by-step explanation:

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3 years ago

Sam and Bethan share £54 in the ratio 5:4 Work out how much each person gets.

jeyben [28]

Answer:

Sam: $30; Bethan: $24

Step-by-step explanation:

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$\frac{x}{54} =\frac{5}{9}$

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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.

Let $\mu$ = mean number of thunder days.

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 One-sample t test statistics 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, test statistics = $\frac{72-92}{\frac{38}{\sqrt{15} } }$ ~ $t_1_4$

= -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 we reject our null hypothesis.

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

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3 years ago