Which equations have a leading coefficient of 3 and a constant term of -2? Check all that apply.

5. A survey of student pizza preferences showed that 43 students preferred cheese, 56 preferred sausage, 39 preferred pepperoni,

cestrela7 [59]

Answer:

P (Cheese) = 0.199, P (Sausage) = 0.259, P (Pepperoni) = 0.181,

P (Supreme) = 0.130, P (Another Kind) = 0.144

and P (Does not like any kind) = 0.088

Step-by-step explanation:

Given:

Number of students who prefer cheese = 43

Number of students who prefer sausage = 56

Number of students who prefer pepperoni = 39

Number of students who prefer supreme = 28

Number of students who prefer another kind = 31

Number of students who did not like any kind = 19

∴ The total number of students surveyed = $43+56+39+28+31+19=216$ The number of students who prefer pizza = $43+56+39+28+31=197$

The probability that a students likes pizza is,

$P(Student\ likes\ pizza)=\frac{No.\ of\ students\ who\ prefer\ pizza}{Total\ no.\ of\ students\ surveyed}$

$=\frac{197}{216} \\=0.912$

The probability that a students does not likes pizza is,

$P(Student\ does\ not\ likes\ pizza)=\frac{No.\ of\ students\ who\ does\ not\ prefer\ pizza}{Total\ no.\ of\ students\ surveyed}$

$=\frac{19}{216} \\=0.088$

The probability distribution of students who prefer different kinds of pizza is:

The probability that a student likes cheese:

$P(A\ Student\ prefers\ cheese)=\frac{No.\ of\ students\ who\ prefer\ cheese}{Total\ no.\ of\ students\ surveyed}$

$=\frac{43}{216}\\=0.199$

The probability that a student likes sausage:

$P(A\ Student\ prefers\ sausage)=\frac{No.\ of\ students\ who\ prefer\ sausage}{Total\ no.\ of\ students\ surveyed}$

$=\frac{56}{216}\\=0.259$

The probability that a student likes pepperoni:

$P(A\ Student\ prefers\ pepperoni)=\frac{No.\ of\ students\ who\ prefer\ pepperoni}{Total\ no.\ of\ students\ surveyed}$

$=\frac{39}{216}\\=0.181$

The probability that a student likes supreme:

$P(A\ Student\ prefers\ supreme)=\frac{No.\ of\ students\ who\ prefer\ supreme}{Total\ no.\ of\ students\ surveyed}$

$=\frac{28}{216}\\=0.130$

The probability that a student likes another kind:

$P(A\ Student\ prefers\ another\ kind)=\frac{No.\ of\ students\ who\ prefer\ another\ kind}{Total\ no.\ of\ students\ surveyed}$

$=\frac{31}{216}\\=0.144$

Thus, the probability distribution table is displayed below:

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3 years ago

Played for 5 days. At the end Of the week she had won 30 games! Every day, she won 5 more games than the day before.

mezya [45]

Ok... what are you trying to ask

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3 years ago

Help please math problem

vichka [17]

The answers are A, C and F

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2 years ago

Independent random samples of vehicles traveling past a given point on an interstate highway have been observed on monday versus

quester [9]

Hi!

To compare this two sets of data, you need to use a t-student test:

You have the following data:

-Monday n1=16; <span>x̄1=59,4 mph; s1=3,7 mph

-Wednesday n2=20; </span>x̄2=56,3 mph; s2=4,4 mph

You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.

$t= \frac{X1-X2}{ \sqrt{ \frac{(n1-1)* s1^{2}+(n2-1)* s2^{2} }{n1+n2-2}} * \sqrt{ \frac{1}{n1}+ \frac{1}{n2}} } =2,2510$

To calculate the degrees of freedom you need to use the following equation:

$df= \frac{ (\frac{ s1^{2}}{n1} + \frac{ s2^{2}}{n2})^{2}}{ \frac{(s1^{2}/n1)^{2}}{n1-1}+ \frac{(s2^{2}/n2)^{2}}{n2-1}}=33,89$ ≈34

The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10

So, as the calculated value is higher than the critical tabulated one, we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.

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84 divided by a number z

Xelga [282]

84/z is your answer.

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