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Lapatulllka [165]

3 years ago

7

The box-and-whisker plot shows how many problems were missed on a test. What is the DIFFERENCE of the interquartile ranges for T

est 1 and Test 2? *

1 answer:

VMariaS [17]3 years ago

3 0

WE NEED THE QUESTION

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Can someone answer the 2 questions in the picture i really need help

Radda [10]

Hi.

The first question is a direct variation — specifically y = 1.8x

In the second one (if my calculations are correct), it's not a direction variation.

Hope this helps :)

Calculations:

y = 1.8(3) = 5.4

y does not equal x, 2x, or -2x.

7 0

3 years ago

Customers arrive at a service facility according to a Poisson process of rate λ customers/hour. Let X(t) be the number of custom

mash [69]

Answer:

Step-by-step explanation:

Given that:

X(t) = be the number of customers that have arrived up to time t.

$W_1,W_2$ ... = the successive arrival times of the customers.

(a)

Then; we can Determine the conditional mean E[W1|X(t)=2] as follows;

$E(W_!|X(t)=2) = \int\limits^t_0 {X} ( \dfrac{d}{dx}P(X(s) \geq 1 |X(t) =2))$

$= 1- P (X(s) \leq 0|X(t) = 2) \\ \\ = 1 - \dfrac{P(X(s) \leq 0 , X(t) =2) }{P(X(t) =2)}$

$= 1 - \dfrac{P(X(s) \leq 0 , 1 \leq X(t)) - X(s) \leq 5 ) }{P(X(t) = 2)}$

$= 1 - \dfrac{P(X(s) \leq 0 ,P((3 \eq X(t)) - X(s) \leq 5 ) }{P(X(t) = 2)}$

Now $P(X(s) \leq 0) = P(X(s) = 0)$

(b) We can Determine the conditional mean E[W3|X(t)=5] as follows;

$E(W_1|X(t) =2 ) = \int\limits^t_0 X (\dfrac{d}{dx}P(X(s) \geq 3 |X(t) =5 )) \\ \\ = 1- P (X(s) \leq 2 | X (t) = 5 ) \\ \\ = 1 - \dfrac{P (X(s) \leq 2, X(t) = 5 }{P(X(t) = 5)} \\ \\ = 1 - \dfrac{P (X(s) \LEQ 2, 3 (t) - X(s) \leq 5 )}{P(X(t) = 2)}$

Now; $P (X(s) \leq 2 ) = P(X(s) = 0 ) + P(X(s) = 1) + P(X(s) = 2)$

(c) Determine the conditional probability density function for W2, given that X(t)=5.

So ; the conditional probability density function of $W_2$ given that X(t)=5 is:

$f_{W_2|X(t)=5}}= (W_2|X(t) = 5) \\ \\ =\dfrac{d}{ds}P(W_2 \leq s | X(t) =5 ) \\ \\ = \dfrac{d}{ds}P(X(s) \geq 2 | X(t) = 5)$

7 0

3 years ago

Which model represents 2 . 36 ÷ 4 ?Which model represents 2 . 36 ÷ 4 ? A. Three groups each have three tens and 4 ones. B. There

makvit [3.9K]

Answer: C. There are four groups of three tens and six ones

Step-by-step explanation:

From the options,

A. Three groups each have three tens and 4 ones.

Three tens and 4 ones equal: = (3 × 10) + (4 × 1) = 30 + 4 = 34

Therefore, 34 ÷ 3 is incorrect.

B. There are three groups of five tens and 9 ones in each group.

Five tens and 9 ones will be equal to:

= (5 × 10) + (9 × 1)

= 50 + 9 = 59

Therefore, 59 ÷ 3 is incorrect

C. There are four groups of three tens and six ones.

Three tens and six ones equal:

= (3 × 10) + (6 × 1)

= 30 + 6 = 36

Therefore, 36 ÷ 4 is correct

D. There are three groups of five tens and nine ones in each group.

Five tens and nine ones will be equal to:

= (5 × 10) + (9 × 1)

= 50 + 9 = 59

Therefore, 59 ÷ 3 is incorrect

Therefore, the correct option is C.

5 0

2 years ago

Please answer this correctly

Mila [183]

Answer:

50%

Step-by-step explanation:

not even = 2 nums

pnoteven = 2 ÷ 4 × 100

3 0

3 years ago

Read 2 more answers

Jason’s credit card has an apr of 17.02% and a 30-day billling cycle. the following table details jason’s transactions with that

forsale [732]

The method of computing that would result in a greater finance charge is a. the daily balance method will have a finance charge $1.02 greater than the adjusted balance method.

<h3>What is the Adjusted Balance Method?</h3>

This refers to the method of accounting that makes use of the owed amount of money at the end of a billing cycle to make its computation on an account after the credits are calculated.

Hence, we can see that when comparing the adjusted balance method to the daily balance method that calculates the interest charges at the end of the day, the daily balance method would have a higher finance charge.

Read more about adjusted balance methods here:

brainly.com/question/1808408

<h3>#SPJ4</h3>

5 0

1 year ago