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emmasim [6.3K]
3 years ago
14

Helppppp!9x - 20+ x2 ​

Mathematics
1 answer:
AleksAgata [21]3 years ago
5 0
-31 cause juss do step by step
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Answer: X=2   Y=-6

Step-by-step explanation:

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Tom is going to USA on holiday. he changes £500 to U.S dollar. the exchange rate is £1=$1. 51. how much dollar will he get​
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500 x 1.51

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∴ He will get $755.

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Evan and Dorinda make bracelets for a craft fair.
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98 bracelets

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Find the value of x (11x-15) (5x-13)
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Step-by-step explanation:

(11x - 15) (5x - 13)

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A researcher wishes to see if the average number of sick days a worker takes per year is less than 5. A random sample of 30 work
Lana71 [14]

Answer:

No. At a significance level of 0.01, there is not enough evidence to support the claim that the average number of sick days a worker takes per year is significantly less than 5.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the average number of sick days a worker takes per year is significantly less than 5.

Then, the null and alternative hypothesis are:

H_0: \mu=5\\\\H_a:\mu< 5

The significance level is 0.01.

The sample has a size n=30.

The sample mean is M=4.8.

The standard deviation of the population is known and has a value of σ=1.2.

We can calculate the standard error as:

\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{1.2}{\sqrt{30}}=0.219

Then, we can calculate the z-statistic as:

z=\dfrac{M-\mu}{\sigma_M}=\dfrac{4.8-5}{0.219}=\dfrac{-0.2}{0.219}=-0.913

This test is a left-tailed test, so the P-value for this test is calculated as:

\text{P-value}=P(z

As the P-value (0.181) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.01, there is not enough evidence to support the claim that the average number of sick days a worker takes per year is significantly less than 5.

3 0
3 years ago
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