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Sunny_sXe [5.5K]
3 years ago
10

I need help please & thx

Mathematics
1 answer:
kicyunya [14]3 years ago
3 0

13. y > 2

14. x - 13 < 9;; x < 22

15. x + 86 < 263;; at most 177 pounds

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The perimeter of a rectangular playground is 46m. If the length of the park is 7m, what is the width of the park?
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16m

Step-by-step explanation:

Since the length of the park is 7m, then the other side is also 7m so 46-14 is 32. So 32/2 is 16.

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Suppose x+y=11 and the value of x increases by2 2. if their sum remains the same, what mjst happen to the value of y? justify yo
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Read 2 more answers
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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