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lakkis [162]
2 years ago
12

P(x)=x^4 −2x^3−3x^2+4 What is the remainder when p(x) is divided by (x-3)

Mathematics
1 answer:
babunello [35]2 years ago
3 0

Answer:

A=5

Step-by-step explanation:

Hence the value of a and b are 5

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A drawer contains 6 white socks and 10 black socks. If 2 socks are chosen at random, what is the probability that they will have
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|\Omega|=C(16,2)=\dfrac{16!}{2!14!}=\dfrac{15\cdot16}{2}=120\\|A|=C(6,2)+C(10,2)=\dfrac{6!}{2!4!}+\dfrac{10!}{2!8!}=\dfrac{5\cdot6}{2}+\dfrac{9\cdot10}{2}=15+45=60\\\\P(A)=\dfrac{60}{120}=\dfrac{1}{2}=50\%

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WILL GIVE BRANLIEST,VOTE, AND RATE!!!!!!!!!!!
kondaur [170]

Step-by-step explanation:

1st digit is 1 number (1)

2nd digit is (3,4,5) so it's 3 numbers

third digit is (6,7,8,9) is it's 4 numbers

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3 years ago
Steven created the scatterplot and trend line below to model the relationship between the number of innings he pitched and the n
-Dominant- [34]

Answer:

The answer is C (7)

Step-by-step explanation:

The graph gives the equation Y=12x, so by doing 87/12 you get 7.25 which is closest to 7 giving you the answer.

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3 years ago
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In a random sample of 85 people attending a workshop, 63 of the respondents said the workshop was useful. What is the sample pro
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Divide the number of respondents saying it was useful by the total number
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8 0
3 years ago
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A national study report indicated that​ 20.9% of Americans were identified as having medical bill financial issues. What if a ne
Alenkasestr [34]

Answer:

The​ p-value for this​ test is 0.22065.

Step-by-step explanation:

We are given that a national study report indicated that​ 20.9% of Americans were identified as having medical bill financial issues.

A news organization randomly sampled 400 Americans from 10 cities and found that 90 reported having such difficulty.

<u><em>Let p = proportion of Americans who were identified as having medical bill financial issues in 10 cities.</em></u>

SO, Null Hypothesis, H_0 : p \leq 20.9%   {means that % of Americans who were identified as having medical bill financial issues in these 10 cities is less than or equal to 20.9%}

Alternate Hypothesis, H_A : p > 20.9%   {means that % of Americans who were identified as having medical bill financial issues in these 10 cities is more than 20.9% and is more severe}

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 400 Americans from 10 cities who were found having such difficulty =  \frac{90}{400} = 0.225 or 22.5%

            n = sample of Americans = 400

So, <u><em>test statistics</em></u>  =  \frac{0.225-0.209}{{\sqrt{\frac{0.225(1-0.225)}{400} } } } }

                               =  0.77

<u></u>

<u>Now, P-value of the test statistics is given by the following formula;</u>

         P-value = P(Z > 0.77) = 1 - P(Z \leq 0.77)

                                            = 1 - 0.77935 = 0.22065

5 0
3 years ago
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