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2 years ago

5

Ben takes a 45 minute test. After answering 10 questions, the timer indicates that one-fifth of the total time is up. How much t

ime did Ben take to answer the 10 questions *

1 answer:

TEA [102]2 years ago

5 0

it took Ben 9 minutes to answer 10 questions

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Find the x- and y-intercepts of the graph of the linear equation −x+8y=4

Andreyy89

First set the equation equal to y by adding x to each side, then divide everything by 8. then set x to zero and solve to find the y - intercept of 1/2 which is just 4/8 reduced. you get this answer because x/8 becoming 0/8 is just 0 leaving you with only 1/2. then reset the equation with x and set y to zero and solve for x to get the x- intercept of -4. you get this by subtracting 1/2 from both sides leaving you -1/2=x/8 and multiply by 8 to get -4=x

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3 years ago

a school sells boxes of cookies each year for a charity fund .the data set shows the number of boxes the 10 families living on w

Thepotemich [5.8K]

0, 1, 1, 2, 2, 2, 2, 3, 3, 5

minimum: 0
maximum: 5
mean: 2.1
median: 2
mode: 2

I think the data have zero skew. Its mean, median, and mode are based on 2. It shows symmetry.

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3 years ago

PLS HELP!!!!!!!!!!!!

seropon [69]

Answer:

72 55/76

Step-by-step explanation:

it's right trust me <3

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2 years ago

Read 2 more answers

An instructor who taught two sections of engineering statistics last term, the first with 20 students and the second with 30, de

Lelechka [254]

Answer:

(a) $P(X=10) = 0.2070$

(b) $P(X\geq 10) = 0.3798$

Step-by-step explanation:

Note that in this problem we have an initial population N = 50, of which 30 fulfill a certain characteristic "m" (belong to the second section). Then, from the population N, a sample of size n = 15 is selected and it is desired to know how many comply with the desired characteristic (second section).

So

Let X be the number of projects in the second section, then X is a discrete random variable that can be modeled by a hypergeometric distribution.

(a)

Therefore, to answer question (a) we use the following equation presented in the attached image:

Where:

$N = 50\\m = 30\\n = 15\\X = 10$

Then:

$P (X = 10) = \frac{\frac{30!}{10!(30-10)!}\frac{20!}{5!(20-5)!}}{\frac{50!}{15!(50-15)!}}\\\\\\P(X=10) = 0.2070$

(b)

For part (b) we have:

$P(X\geq10) = 1-P(X$

$P(X\geq 10) = 1-0.6202 \\\\P(X\geq 10) = 0.3798$

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3 years ago

Hey Guys I have a question can yall help me with the quiz

LiRa [457]

Answer:

sure

Step-by-step explanation:

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3 years ago