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stira [4]
3 years ago
6

A pizza place offers the following toppings: pepperoni, sausage, onions. green peppers, and mushrooms.

Mathematics
1 answer:
Harman [31]3 years ago
5 0

Answer:

10

Step-by-step explanation:

You do 5C3, which is 5!/((5-3)!3!) -->

5*4*3*2*1/(2*1*3*2*1) =

10

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3 0
2 years ago
Given f(x) = 4x^2 + 19x - 5 and g(x) = 4x^2 - x what is (f/g)(x)?
Free_Kalibri [48]

( \frac{f}{g} )(x) =  \frac{4 {x}^{2}  + 19x - 5}{4 {x}^{2}  - x}  =  \frac{(4x - 1)(x + 5)}{x(4x - 1)}  =  \frac{x + 5}{x}

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3 years ago
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In a restaurant, the proportion of people who order coffee with their dinner is p. A simple random sample of 144 patrons of the
mote1985 [20]
<h2>Answer with explanation:</h2>

Given : In a restaurant, the proportion of people who order coffee with their dinner is p.

Sample size : n= 144

x= 120

\hat{p}=\dfrac{x}{n}=\dfrac{120}{144}=0.83333333\approx0.8333

The null and the alternative hypotheses if you want to test if p is greater than or equal to 0.85 will be :-

Null hypothesis : H_0: p\geq0.85   [ it takes equality (=, ≤, ≥) ]

Alternative hypothesis : H_1: p  [its exactly opposite of null hypothesis]

∵Alternative hypothesis is left tailed, so the test is a left tailed test.

Test statistic : z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}

z=\dfrac{0.83-0.85}{\sqrt{\dfrac{0.85(1-0.85)}{144}}}\\\\=-0.561232257678\approx-0.56

Using z-vale table ,

Critical value for 0.05 significance ( left-tailed test)=-1.645

Since the calculated value of test statistic is greater than the critical value , so we failed to reject the null hypothesis.

Conclusion : We have enough evidence to support the claim that p is greater than or equal to 0.85.

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3 years ago
48 gallons in 14 mins simplified
zvonat [6]
48/14 so you just simplify that for gallons/minutes. something that goes into both of those numbers is 2 so you would have 24/7 when you divide by 2. that’s your answer.
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