MATHEMATICS COLLEGE

In a survey of 269 college students, it is found that69 like brussels sprouts,
90 like broccoli,
59 like cauliflower,
28 like both Brussels sprouts and broccoli,
20 like both Brussels sprouts and cauliflower,
24 like both broccoli and cauliflower, and
10 of the students like all three vegetables.

a) How many of the 269 college students do not like any of these three vegetables?

b) How many like broccoli only?

c) How many like broccoli AND cauliflower but not Brussels sprouts?

d) How many like neither Brussels sprouts nor cauliflower?

Answers

Answer 1

Answer:

Answer: a) 83, b) 28, c) 14, d) 28.

Step-by-step explanation:

Since we have given that

n(B) = 69

n(Br)=90

n(C)=59

n(B∩Br)=28

n(B∩C)=20

n(Br∩C)=24

n(B∩Br∩C)=10

a) How many of the 269 college students do not like any of these three vegetables?

n(B∪Br∪C)=n(B)+n(Br)+n(C)-n(B∩Br)-n(B∩C)-n(Br∩C)+n(B∩Br∩C)

n(B∪Br∪C)= $69+90+59-28-20-24+10=156$

So, n(B∪Br∪C)'=269-n(B∪Br∪C)=269-156=83

b) How many like broccoli only?

n(only Br)=n(Br) -(n(B∩Br)+n(Br∩C)+n(B∩Br∩C))

n(only Br)= $90-(28+24+10)=28$

c) How many like broccoli AND cauliflower but not Brussels sprouts?

n(Br∩C-B)=n(Br∩C)-n(B∩Br∩C)

n(Br∩C-B)= $24-10=14$

d) How many like neither Brussels sprouts nor cauliflower?

n(B'∪C')=n(only Br)= 28

Hence, a) 83, b) 28, c) 14, d) 28.

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Find the distance between the two points

Answers

I got 8.1 units
.
I used the Pythagorean Theorem to find my answer. I plugged in 7 for the long side and 4 for the short side and then solved.
.
Hope this helps :)

3. Kirk bought a bag of candy and took 10pieces. He split the rest evenly among 12
friends. Each friend received 5 pieces. Letc
represent the number of pieces in a bag.
Equation:
Solve it to find how many pieces of candy were in the bag.
Type here
Show your work
Write and solve the equation

Answers

70, Multiply 5 pieces by 12 friends and you get 60, then add the 10 pieces he originally took out, and you get 70

5x5x5x5= ______ to the power of _____

Answers

Answer:

5 4

Step-by-step explanation:

its fristly 5

ok then bb

Answer:

5x5x5x5=625

Step-by-step explanation:

PLEASE HELPP MEE ASAPP!!

Answers

Answer:

Step-by-step explanation:

each zero of the function will have a factor of (x - x₀)

h(x) = a(x + 3)(x + 2)(x - 1)

h(x) = a(x + 3)(x² + x - 2)

h(x) = a(x³ + 4x² + x - 6)

or the third option works if a = 1

however this equation gives us the points (0, -6) and (-1. -4), so "a" must be -2

h(x) = -2x³ - 8x² - 2x + 12

to fit ALL of the given points as it fits the three zeros and also h(0) and h(-1) so I guess that is why the given group is a partial set of solution sets

What is the approximate circumference pf the circle shown below?

Answers

Answer:

A: 20pi

Step-by-step explanation:

c=2(pi)r

c=2(pi)(10)

c=20(pi)

c=62.8

The mean SAT score in mathematics, M, is 600. The standard deviation of these scores is 48. A special preparation course claims that its graduates will score higher, on average, than the mean score 600. A random sample of 70 students completed the course, and their mean SAT score in mathematics was 613. a) At the 0.05 level of significance, can we conclude that the preparation course does what it claims? Assume that the standard deviation of the scores of course graduates is also 48.

Answers

Answer:

Step-by-step explanation:

The mean SAT score is $\mu=600$ , we are going to call it \mu since it's the "true" mean

The standard deviation (we are going to call it $\sigma$ ) is

$\sigma=48$

Next they draw a random sample of n=70 students, and they got a mean score (denoted by $\bar x$ ) of $\bar x=613$

The test then boils down to the question if the score of 613 obtained by the students in the sample is statistically bigger that the "true" mean of 600.

- So the Null Hypothesis $H_0:\bar x \geq \mu$

- The alternative would be then the opposite $H_0:\bar x < \mu$

The test statistic for this type of test takes the form

$t=\frac{| \mu -\bar x |} {\sigma/√(n)}$

and this test statistic follows a normal distribution. This last part is quite important because it will tell us where to look for the critical value. The problem ask for a 0.05 significance level. Looking at the normal distribution table, the critical value that leaves .05% in the upper tail is 1.645.

With this we can then replace the values in the test statistic and compare it to the critical value of 1.645.

$t=\frac{| \mu -\bar x |} {\sigma/√(n)}\n\n= (| 600-613 |)/(48/\sqrt(70)}\n\n= (| 13 |)/(48/8.367)\n\n= (| 13 |)/(5.737)\n\n=2.266\n$

since 2.266>1.645 we can reject the null hypothesis.

Answer:

The null hypothesis is that the SAT score is not significantly different for the course graduates.

Alternate hypothesis: there is a significant difference between the SAT score achieved by the course graduates as compared to the non-graduates.

Apply the t-test. The Test Statistic value comes out to be t = 1.738 and the p-value = 0.0844

Since the p-value is larger than 0.05, the evidence is weak and we fail to reject eh null hypothesis.

Hope that answers the question, have a great day!

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