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Liono4ka [1.6K]

2 years ago

11

The chart shows how many people have signed up to go on a field trip each day. 62 students are allowed to go on the field trip.

On which day would you expect that number to be reached?A) 8B) 9C) 10D) 11

1 answer:

Alexeev081 [22]2 years ago

7 0

Answer:

C

I hope it helps.

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Please help steps thx

ExtremeBDS [4]

Answer:

240 cubic metres

Step-by-step explanation:

I'm not really sure for Q17 but I do know Q18.

Q18. Volume of cuboid = Length×Breadth×Height

=12×4×5

=240 cubic metres

hope it helps!!!

8 0

2 years ago

0.27 x 0.03 PLEASE EXPLAIN THE ANSWER! HOW DID YOU GET THE SUM??<br><br><br> I NEED HELP PLEASE!

kow [346]

It would be 0.008 because you line up the decimals and multiply

8 0

3 years ago

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In which equation x is = to 180<br> (A) x/49=7 (B) x/12-7 =13 (C) x/3-15=45

topjm [15]

The answers is C. because when you do 180/3-15 you get 45 as the answer.

7 0

3 years ago

Change 3/4 into its equivalent decimal form

svet-max [94.6K]

That would be 0.75:) hope this helps!

6 0

3 years ago

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Consider a large shipment of 400 refrigerators, of which 40 have defective compressors. If X is the number among 15 randomly sel

Vlad [161]

Answer:

The required probability is 0.94

Step-by-step explanation:

Consider the provided information.

There are 400 refrigerators, of which 40 have defective compressors.

Therefore N = 400 and X = 40

The probability of defective compressors is:

$\frac{40}{400}=0.10$

It is given that If X is the number among 15 randomly selected refrigerators that have defective compressors,

That means n=15

Apply the probability density function.

$P(X=x)=^nC_xp^x(1-p)^{n-x}$

We need to find P(X ≤ 3)

$P(X\leq3) =P(X=0)+P(X=1)+P(X=2)+P(X=3)\\P(X\leq3) =\frac{15!}{15!}(0.1)^0(1-0.1)^{15}+\frac{15!}{14!}(0.1)^1(1-0.1)^{14}+\frac{15!}{13!2!}(0.1)^2(1-0.1)^{13}+\frac{15!}{12!3!}(0.1)^3(1-0.1)^{12}\\$

$P(X\leq3) =0.944444369992\approx 0.94$

Hence, the required probability is 0.94

4 0

2 years ago