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2 years ago

Find the sum of the average monthly rainfall. this is on i ready okay so please help.

1 answer:

6 0

Based on the average monthly rainfall amounts as shown in the graph, the sum of the average monthly rainfall 11/12 inches.

<h3>What is the sum of the average monthly rainfall?</h3><h3 />

The average can be found by adding up the monthly rainfall and dividing by 12 months.

Solving gives:

= (1/8 + 1/8 + 1/8 + 2/8+ 2/8 + 5/8 + 7/8 + 9/8 + 13/8 + 15/8 + 16/8 + 16/8) / 12

= (88/8) / 12

= 11/12 inches

Find out more on average rainfall at brainly.com/question/23871385

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Delicious77 [7]

Answer:

The answer is equivalent

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2 years ago

NEED HELP ASAP !!!!!!!!

Papessa [141]

Answer: Stephon can buy total 4 chips bag and 2 candy bar

Step-by-step explanation:Given as :

The money which Stephon has in his pocket = $ 12

The cost of one chips bag = $ 2.25

The cost of one candy bar = $ 1.50

The total number of items he can buy = 6

So, Let the number of chips bag he can buy = x

∴ The number of candy bar he can buy = 6 - x

Now, according to question

The number of chips bag×The cost of one chips bag + The number of candy bar×The cost of one candy bar = $ 12

Or, x × $ 2.25 + (6-x) × $ 1.50 = $ 12

Or, x × $ 2.25 + 9 - x × $ 1.50 = $ 12

or, x × $ 0.75 = 12 - 9

or, x = = 4

The number of chips bag he can buy = x = 4

And The number of candy bar he can buy = 6 - x = 6 - 4 = 2

Hence Stephon can buy total 4 chips bag and 2 candy bar .

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2 years ago

Subtract 5x^2 – 2x – 5 from – 2x^2<br> - 5x - 7

user100 [1]

Answer:

$-7x^{2} -3x-2$

Step-by-step explanation:

$(-2x^2-5x-7) - (5x^2-2x-5)\\\\-2x^2 - 5x^2 = -7 x^2\\-5x - -2x = -3x\\-7 - -5 = -2 \\Result: -7x^2-3x-2$

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2 years ago

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Discrete R.V. Assume a student shows up for EGR 280 completely unprepared and the instructor gives a pop quiz. Assume there are

Tom [10]

Answer:

a) p=0.2

b) probability of passing is 0.01696

c) The expected value of correct questions is 1.2

Step-by-step explanation:

a) Since each question has 5 options, all of them equally likely, and only one correct answer, then the probability of having a correct answer is 1/5 = 0.2.

b) Let X be the number of correct answers. We will model this situation by considering X as a binomial random variable with a success probability of p=0.2 and having n=6 samples. We have the following for k=0,1,2,3,4,5,6

$P(X=k) = \binom{n}{k}p^{k}(1-p)^{n-k} = \binom{6}{k}0.2^{k}(0.8)^{6-k}$ .

Recall that $\binom{n}{k}= \frac{n!}{k!(n-k)!}$ In this case, the student passes if X is at least four correct questions, then

$P(X\geq 4) = P(X=4)+P(X=5)+P(X=6)=\binom{6}{4}0.2^{4}(0.8)^{6-4}+\binom{6}{5}0.2^{5}(0.8)^{6-5}+\binom{6}{6}0.2^{6}(0.8)^{6-6}= 0.01696$

c)The expected value of a binomial random variable with parameters n and p is $E[X] = np$ . IN our case, n=6 and p =0.2. Then the expected value of correct answers is $6\cdot 0.2 = 1.2$

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3 years ago

How do I solve N(1/5) = 2/15

dem82 [27]

N(1/5) = (2/15)
divide both sides by (1/5)
n = (2/15)/(1/5)
to divide fractions, cross multiply, e.g. (2*5)/(15*1) = 10/15
n = 10/15 = 2/3 or 0.666666.

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3 years ago

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