A grocery store manager claims that 75% of shoppers purchase bananas as least once a month. Technology was used to simulate choosing 150 SRSs of size n = 100 from a population of shoppers where 75% buy bananas. The dotplot shows pˆ = the sample proportion of shoppers who bought bananas in the past month. A random sample of 100 shoppers from the store were selected and 64 bought bananas in the past month. Does this sample provide evidence that the grocery store manager overstated the true proportion? Justify your answer.

P-value = 0.0333

At 5% level of significance;

0.0333 < 0.05

Therefore, we reject null hypothesis H₀ at 5% level of significance,

We conclude that proportion of shoppers who bought bananas at least once in the past month is overstated

Step-by-step explanation:

 Given the data in the question;

To test whether population proportion p is overstated;

Null hypothesis H₀ : p = (75%) = 0.75

Alternative hypothesis H₁ : = < (75%) < 0.75

now, sample proportion p" = 64 / 100 = 0.64

from the dot plot below, we will determine the p-value for test { P(p" < 0.64)}

so, the number of times p"<0.64 in 150 simulations is 5

Hence; P(p" < 0.64 ) = 5 / 150 = 0.0333

P-value = 0.0333

At 5% level of significance;

0.0333 < 0.05

Therefore, we reject null hypothesis H₀ at 5% level of significance,

We conclude that proportion of shoppers who bought bananas at least once in the past month is overstated

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

cretdit to source

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step-by-step explanation:

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