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3 years ago

How do I solve this?

Mathematics

1 answer:

vladimir1956 [14]3 years ago

4 0

Answer:

3p³ + 2p² – 3p – 11

Step-by-step explanation:

From the question given above, the following data were obtained:

Side 1 (S₁) = –1(p + 5)

Side 2 (S₂) = 2(p² – 3)

Side 3 (S₃) = 3p³ – 2p

Perimeter (P) =?

The perimeter of the triangle can be obtained as follow

P = S₁ + S₂ + S₃

P = –1(p + 5) + 2(p² – 3) + 3p³ – 2p

Clear bracket

P = –p – 5 + 2p² – 6 + 3p³ – 2p

Rearrange

P = 3p³ + 2p² – 2p – p – 6 – 5

P = 3p³ + 2p² – 3p – 11

Therefore, the perimeter of the triangle is 3p³ + 2p² – 3p – 11

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Select all the exspressions that are all equivalent to 20 percent of 150

Dmitriy789 [7]

Note: It seems you may have unintentionally missed adding the answer choices. Thus, I am solving your question in general to give you the idea of how the percentage works, which anyways would solve your query.

Answer:

Please check the explanation.

Step-by-step explanation:

Given that we have to determine the expressions which are equivalent to 20 percent of 150.

First, we need to determine what actually 20 percent of 150 really brings.

i.e

20% of 150 = 20/100 × 150

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20% of 150 = 30

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3 years ago

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4 years ago

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2 years ago

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3 years ago

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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.

Let p = proportion of all cars in San Francisco who are blue

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 One-sample z proportion statistics;

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, test statistics = $\frac{0.235-0.25}{{\sqrt{\frac{0.235(1-0.235)}{600} } } } }$

= -0.866

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.

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

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3 years ago