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denis-greek [22]

3 years ago

12

Russell runs 9/10 mile in five minutes. How many miles does he run in one minute

2 answers:

shepuryov [24]3 years ago

8 0

9/50 of a mile, you have to divide 9/10 by 5 to find the number of mile(s) he ran in a minute.

dem82 [27]3 years ago

3 0

9/10=5mins
times both sides by 1/5
9/50=1min

he can run 9/50 miles per minute

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If y = kx, what is the value of k if y = 20 and x = 0.4? A) 5 B) 8 C) 9.6 D) 50

Margaret [11]

Answer: D, 50
y = kx <span>y = 20 and x = 0.4
</span>
What you will do: You will plug in your y and your x.
Your equation should look like: 20=k(0.4)
Next: What you will do is divide each side of your equal sign by 0.4 so you will get your k alone.
You would get: k=50

HOPE THIS HELPS! ^_^

7 0

3 years ago

In the data set shown below, what is the value of the quartiles? {42, 43, 44, 44, 48, 49, 50} A. Q1 = 43.5; Q2 = 44; Q3 = 49 B.

ANTONII [103]

Answer:

C

Step-by-step explanation:

I don't actually know how to explain it but,

Q1=43.5

Q2=44

Q3=48.5

Sorry, I can't help more.

3 0

3 years ago

How many words of five letters can be created that start with the letter T, contain 2 other consonats and end with 2 vowels? Not

Jet001 [13]

If we're only counting 5 vowels (A, E, I, O, U) and 20 consonants (everything else, minus T), then there are

$\dbinom52=\dfrac{5!}{2!(5-2)!}=10$

ways of picking the vowels, and

$\dbinom{20}2=\dfrac{20!}{2!(20-2)!}=190$

ways of picking the consonants.

We want the word to start with T, and we'll allow any arrangement of the other 4 letters, so that the total number of words is

$4!\dbinom52\dbinom{20}2=\boxed{45,600}$

Keep in mind that this means words like TRIES and TIRES are treated as different.

8 0

3 years ago

Find the probability of having 2,3, or 4 successes in five trials of a binomial experiment in which the probability of success i

Georgia [21]

Answer:

65.3%

Step-by-step explanation:

Binomial distribution PMF below (along with a description)

the probabilty of getting 2, 3, or 4 successes in 5 trials is just

p(2)+p(3)+p(4)

p(2)= 5C2*.4²*(1-.4)³ = .3456

p(3)= 5C3*.4³*(1-.4)²= .2304

p(4)= 5C4*.4⁴*(1-.4)=.0768

.3456+.2304+.0768= .6528

6 0

2 years ago

The difference of two numbers is 9 and the sum of the numbers is 55. what are the numbers?

yKpoI14uk [10]

The two numbers I think is 3&6 because this is a trick question

3 0

3 years ago