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3 years ago

Can someone help me with 8 please

Mathematics

1 answer:

Dmitry [639]3 years ago

4 0

Answer:

96,144,192,240,288

Step-by-step explanation:

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Write each fractio in simplest form.4/10

hjlf

4/10 would be 2/5 in simplest form.

I hope this helps! (:

5 0

3 years ago

Read 2 more answers

Using the order of operations, what would be the first step:<br><br> 4−8÷1−3[3+(7−5)2]

jeka94

Answer:

(7-5)

Step-by-step explanation:

According to PEMDAS, parenthesis go first. But there are two sets of parenthesis so the one inside the other goes first.

Next would be [3+(2*2)]

Then, [3+4]

Then, -8/1

Then, -3(7)

So the question would be 4-8-21

Which gives you -25 as an answer.

I realize that you weren't asking for the whole thing but maybe there are follow up questions.

3 0

3 years ago

An educator wants to determine whether a new curriculum significantly improves standardized test scores for fourth grade student

vesna_86 [32]

Answer:

Independent Sampling

Step-by-step explanation:

There are two scenarios for independent sampling .

Testing the mean we get the differences between samples from each population. When both samples are randomly inferences about the populations. can get.

Independent sampling are sample that are selected randomly. Observation does not depend upon value.Many analysis assume that sample are independent.

In this statement 90 dash are divides into two groups Group 1 and Group 2 . Both are standardized that mean both are randomly selected. Means are observed. Observation doesn't depend upon value. So this style of sampling is independent Sample.

5 0

3 years ago

What is the answer to this problem?

eimsori [14]

Answer:

A=3

Step-by-step explanation:

srry if I'm worng. hop it helped

6 0

3 years ago

A sample of 4 different calculators is randomly selected from a group containing 18 that are defective and 35 that have no defec

Sphinxa [80]

<span>1.

P(</span>at least one of the calculators is defective)=

1- P(none of the selected calculators is defective).

2.

P(none of the selected calculators is defective)

=n(ways of selecting 4 non-defective calculators)/n(total selections of 4)

3.

selecting 4 non-defective calculators can be done in C(35, 4) many ways,

where $C(35, 4)= \frac{35!}{4!31!}= \frac{35*34*33*32*31!}{4!*31!}= \frac{35*34*33*32}{4!}= \frac{35*34*33*32}{4*3*2*1}= 52,360$

while, the total number selections of 4 out of 18+35=53 calculators can be done in C(53, 4) many ways,

$C(53, 4)= \frac{53!}{4!*49!}= \frac{53*52*51*50}{4*3*2*1}= 292,825$

4. so, P(none of the selected calculators is defective)= $\frac{52,360}{292,825} =0.18$

5. P(at least one of the calculators is defective)=

1- P(none of the selected calculators is defective)=1-0.18=0.82

Answer:0.82

7 0

3 years ago