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abruzzese [7]
3 years ago
11

An art supply store sells posters for $9 each and picture frames for $15 each. Write an expression for the total cost of 6 poste

rs and 6 frames.
Mathematics
1 answer:
slega [8]3 years ago
6 0
Poster=6$
Frame=15$

9p+15f=X
9(6)+15(6)=X
54+90=X
X=144

Total cost is 144$
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Find AB if A(0,3) and B(2,7)<br><br> AB= ?
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Factor completely.
tatyana61 [14]

Answer:

A.) (7t³ + 2k^4)(7t³ - 2k^4)

Step-by-step explanation:

Factor the following:

49 t^6 - 4 k^8

49 t^6 - 4 k^8 = (7 t^3)^2 - (2 k^4)^2:

(7 t^3)^2 - (2 k^4)^2

Factor the difference of two squares. (7 t^3)^2 - (2 k^4)^2 = (7 t^3 - 2 k^4) (7 t^3 + 2 k^4):

Answer:  (7 t^3 - 2 k^4) (7 t^3 + 2 k^4)

3 0
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Read 2 more answers
The opposite of twelve percent of a number​
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5 0
3 years ago
A local car dealer claims that 25% of all cars in San Francisco are blue. You take a random sample of 600 cars in San Francisco
sammy [17]

Answer:

No, we can't reject the dealer's claim with a significance level of 0.05.

Step-by-step explanation:

We are given that a local car dealer claims that 25% of all cars in San Francisco are blue.

You take a random sample of 600 cars in San Francisco and find that 141 are blue.

<u><em>Let p = proportion of all cars in San Francisco who are blue</em></u>

SO, Null Hypothesis, H_0 : p = 25%   {means that 25% of all cars in San Francisco are blue}

Alternate Hypothesis, H_A : p \neq 25%   {means that % of all cars in San Francisco who are blue is different from 25%}

The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;

                                  T.S.  = \frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }  ~ N(0,1)

where, \hat p = sample proportion of 600 cars in San Francisco who are blue =   \frac{141}{600} = 0.235

            n = sample of cars = 600

So, <u><em>test statistics</em></u>  =  \frac{0.235-0.25}{{\sqrt{\frac{0.235(1-0.235)}{600} } } } }

                               =  -0.866

<em>Now at 0.05 significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics lies within the range of critical values of z so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.</em>

Therefore, we conclude that 25% of all cars in San Francisco are blue which means the dealer's claim was correct.

4 0
3 years ago
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