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Arturiano [62]
3 years ago
14

Write an expression to represent: Four less than the product of one and a number xxx

Mathematics
2 answers:
lyudmila [28]3 years ago
5 0

Answer:

(1xxx)-4

Step-by-step explanation:

Since 1 and xxx are multiplied, they produce 1xxx and eventually 4 is subtracted so 1xxx-4 and I put the brackets just to make it clear about what is going on in the expression

rusak2 [61]3 years ago
4 0

Answer:

xxx-4

Step-by-step explanation:

(1×xxx)-4

yietisyksykd

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Number of algae in 7 days = 70.355

Step-by-step explanation:

Given:

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Rate of increasing = 5% per day = 0.05

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Computation:

Number\ of\ algae\ in\ 7\ days = Algae\ in\ beginning(1+Rate\ of\ increasing)^{Number\ of\ days}\\\\Number\ of\ algae\ in\ 7\ days = 50(1+0.05)^7\\\\Number\ of\ algae\ in\ 7\ days = 70.355

Number of algae in 7 days = 70.355

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The conditions for use of the normal model to represent the distribution of sample proportion are not met. He should increase the sample size until the conditions are met.

If the test is done anyway, the null hypothesis failed to be rejected.

The conclusion is taht there is not enough evidence to support the claim that the percentage is lower in this district.

Step-by-step explanation:

The conditions for use of the normal model to represent the distribution of sample proportion are not met, as the affirmative responses are less than 10.

np=80*0.1=8

If the test of hypothesis is done as if the conditiones were met, we know that the claim is that  the percentage is lower in this district.

Then, the null and alternative hypothesis are:

H_0: \pi=0.173\\\\H_a:\pi

The significance level is 0.05.

The sample has a size n=80.

The sample proportion is p=0.1.

 

The standard error of the proportion is:

\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.173*0.827}{80}}\\\\\\ \sigma_p=\sqrt{0.001788}=0.042

Then, we can calculate the z-statistic as:

z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.1-0.173+0.5/80}{0.042}=\dfrac{-0.067}{0.042}=-1.578

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

P-value=P(z

As the P-value (0.057) is greater than the significance level (0.05), the effect is  not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that  the percentage is lower in this district.

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