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3 years ago

(08.03)Jasper is using the following data samples to make a claim about the house values in his neighborhood:

1 answer:

7 0

I think its D (or #4 - He Should use the mean because there are no outliers that affect the mean) because the Median would be $175,000 but the Mean would be $638,400. It's honestly almost up to you.

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Contruct 3 equation starting with -2

olga_2 [115]

Answer:

1. -2, 2. -23, 3. 25

Step-by-step explanation:

1. -2+5x8/-12= step 1: add -2+5

-2+5=3 step 2: 3(8)

3x8= 24 step 3: 24/-12

24/-12= -2

2. -2(9-4)+3=? step 1: combine like terms -2(9) and -2(-4)

-18-8+3=? step 2: combine like terms

-26+3=? step 3: combine like terms

-23=?

3. 2x + 130=180 step 1: -130 on both sides

2x = 50 step 2: /2 on both sides

x = 25

7 0

2 years ago

Read 2 more answers

HELP ASAP PLSS 10 POINTS

Monica [59]

The scale factor is 2.5

let me know if you have any ?s

5 0

3 years ago

Read 2 more answers

What is my fav color/ what is statistical and non statistical

frez [133]

Answer:

changed my mind i say pink and rose gold but for me its Blue

and

statistical is relating to the use of statistics.

and

non statistical is not of, relating to, based on, or employing the principles of statistics

If brainiest is earned its greatly appreciated.

3 0

3 years ago

Please answer quickly 40 points

Elis [28]

Answer:

2/5 hour is closest to 0 1/2 hours, 5/8 hour is closest to 1/2 hour. Therefore, the best estimate for the total time Maria spent with her sister is close to 1 hour

Step-by-step explanation:

We can represent the time Maria spends playing the game as 0.4, since 2/5 as a fraction is 0.4. This is closer to 1/2, which is 0.5 then it is to 0, because 4 is closer to 5 then 0. Then, we can represent the time Maria reads the book as 0.625, which is closer to 1/2 then it is to 1. Now we can add the estimated times. $1/2+1/2=1$ hour

8 0

2 years ago

A bag contains 6 red apples and 5 yellow apples. 3 apples are selected at random. Find the probability of selecting 1 red apple

tester [92]

Answer: The required probability of selecting 1 red apple and 2 yellow apples is 36.36%.

Step-by-step explanation: We are given that a bag contains 6 red apples and 5 yellow apples out of which 3 apples are selected at random.

We are to find the probability of selecting 1 red apple and 2 yellow apples.

Let S denote the sample space for selecting 3 apples from the bag and let A denote the event of selecting 1 red apple and 2 yellow apples.

Then, we have

$n(S)=^{6+5}C_3=^{11}C_3=\dfrac{11!}{3!(11-3)!}=\dfrac{11\times10\times9\times8!}{3\times2\times1\times8!}=165,\\\\\\n(A)\\\\\\=^6C_1\times^5C_2\\\\\\=\dfrac{6!}{1!(6-1)!}\times\dfrac{5!}{2!(5-2)!}\\\\\\=\dfrac{6\times5!}{1\times5!}\times\dfrac{5\times4\times3!}{2\times1\times3!}\\\\\\=6\times5\times2\\\\=60.$

Therefore, the probability of event A is given by

$P(A)=\dfrac{n(A)}{n(S)}=\dfrac{60}{165}=\dfrac{4}{11}\times100\%=36.36\%.$

Thus, the required probability of selecting 1 red apple and 2 yellow apples is 36.36%.

4 0

3 years ago