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Katen [24]
4 years ago
6

Please help! Algebra 1 math.

Mathematics
1 answer:
olya-2409 [2.1K]4 years ago
4 0
A.   The data in the table represent a function. y=x/2. This is seen as for every value of x in the table, y=x/2(half of it). You can show this further by trying this fact on each value.

B.   For the original function, when x =8, y=x/2=4
For the function given in this part, when x=8, y=3(8)-10
                                                                 y=24-10                                                                                                            y=14
     This means the relation given in this part has a greater value when x=8

C. When x=80, f(x)=3(80)-10
                           =240-10
                           =230 
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7 0
3 years ago
A sample of 900900 computer chips revealed that 49%49% of the chips do not fail in the first 10001000 hours of their use. The co
dedylja [7]

Answer:

The proportion of chips that do not fail in the first 1000 hours of their use is 52%.

Step-by-step explanation:

The claim made by the company is that 52% of the chips do not fail in the first 1000 hours of their use.

A quality control manager wants to test the claim.

A one-proportion <em>z</em>-test can be used to determine whether the proportion of chips do not fail in the first 1000 hours of their use is 52% or not.

The hypothesis can be defined as:

<em>H₀</em>: The proportion of chips do not fail in the first 1000 hours of their use is 52%, i.e. <em>p</em> = 0.52.

<em>Hₐ</em>: The proportion of chips do not fail in the first 1000 hours of their use is different from 52%, i.e. <em>p</em> ≠ 0.52.

The information provided is:

\hat p=0.49\\n=900\\\alpha =0.02

The test statistic value is:

z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.49-0.52}{\sqrt{\frac{0.52(1-0.52)}{900}}}=-1.80

The test statistic value is -1.80.

Decision rule:

If the <em>p-</em>value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

Compute the <em>p</em>-value as follows:

p-value=2\times P(Z

*Use a <em>z</em>-table.

The <em>p</em>-value = 0.07186 > <em>α</em> = 0.02.

The null hypothesis was failed to be rejected at 2% level of significance.

<u>Conclusion</u>:

The proportion of chips that do not fail in the first 1000 hours of their use is 52%.

8 0
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Step-by-step explanation:

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3/4 of the pot is left over
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