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3 years ago

If you roll a dice 48 times how many times would you predict not to roll a 6?

2 answers:

4 0

Hi there!

Since there is a 1/6 chance of rolling a 6, we can just divide 48 by 6 and subtract it from 48:

48 / 6 = 8
48 - 8 = 40

So, the amount of times you would not roll a 6 is 40 times.

Hope this helps!

MakcuM [25]3 years ago

3 0

Expected value (1/6) (120)=20

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A new Community Center is being built in Oak Valley. The perimeter of the rectangular playing field is 382 yards. The length of

vesna_86 [32]

Answer:

The width is 50 yards and the length is 141 yards.

Step-by-step explanation:

Let's call: L the length of the field and W the width of the field.

From the sentence, the perimeter of the rectangular playing field is 382 yards we can formulate the following equation:

2L + 2W = 382

Because the perimeter of a rectangle is the sum of two times the length with two times the width.

Then, from the sentence, the length of the field is 9 yards less than triple the width, we can formulate the following equation:

L = 3W - 9

So, replacing this last equation on the first one and solving for W, we get:

2L + 2W = 382

2(3W - 9) + 2W = 382

6W -18 +2W = 382

8W - 18 = 382

8W = 382 + 18

8W = 400

W = 400/8

W = 50

Replacing W by 50 on the following equation, we get:

L = 3W - 9

L = 3(50) - 9

L = 141

So, the width of the rectangular field is 50 yards and the length is 141 yards.

3 0

3 years ago

The body weight of a healthy 3 month-old colt should be about μ = 60 kg. (Source: The Merck Veterinary Manual, a standard refere

tatuchka [14]

Answer:

The null hypothesis will be $H_o : \mu = 60 \ kg$

The alternative hypothesis will be $H_a : \mu < 60$

The alternative hypothesis will be $H_a : \mu > 60$

The alternative hypothesis will be $H_a : \mu \ne 60$

left; right; both

Step-by-step explanation:

Considering question a

The body weight of the colt is $\mu = 60 \ kg$

The null hypothesis will be $H_o : \mu = 60 \ kg$

Considering question b

Given that we want to test the claim that the average weight of a wild Nevada colt is less than 60 kg , then

The alternative hypothesis will be $H_a : \mu < 60$

Considering question c

Given that we want to test the claim that the average weight of such a wild colt is greater than 60 kg , then

The alternative hypothesis will be $H_a : \mu > 60$

Considering question d

Given that we want to test the claim that the average weight of such a wild colt is different from 60 kg , then

The alternative hypothesis will be $H_a : \mu \ne 60$

Considering question e

For question b , the corresponding p-value will be on the left because the average weight is less than 60 kg

For question c , the corresponding p-value will be on the right because the average weight is more than 60 kg

For question d , the corresponding p-value will be on both because the average weight is different from 60 kg

5 0

2 years ago

(-w^3 + 8w^2 – 3w) + (-8w^2 + w + 3) =

kherson [118]

The expression simplified is -w^3-2w+3

I hope this helped

4 0

3 years ago

What value of X makes equation below true 2x + 9 = 4x - 1

ale4655 [162]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

A. 3<br> B. 4<br> C. 130<br> D. 35

Setler79 [48]

Answer:

a) 3

Step-by-step explanation:

3+10= 13

hope this helped :)

7 0

3 years ago

Read 2 more answers