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3 years ago

10

Drag each scenario to show whether the final result will be greater than the original value, less than the original value, or th

e same as the original value.

1 answer:

cestrela7 [59]3 years ago

8 0

Answer:

I have my correct answers in the photo so just read from that :) <3

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PLEASE PLEASE HELP ITS MY FINAL GRADE PLEASE PLEASE

Sati [7]

Answer:

See below

Step-by-step explanation:

Given there are 56 M&M's:

Expected Value of Blue = 56(0.24) = 13.44
Expected Value of Brown = 56(0.14) = 7.84
Expected Value of Green = 56(0.16) = 8.96
Expected Value of Orange = 56(0.20) = 11.20
Expected Value of Red = 56(0.13) = 7.28
Expected Value of Yellow = 56(0.14) = 7.84

Remember to use $np$ to calculate expected value! This will especially be helpful if you are conducting a chi-squared test of independence.

6 0

2 years ago

3(2x - 1) + 7 = -44 solve for x

Jet001 [13]

Answer:

I love algebra anyways

The ans is in the picture with the steps how i got it

(hope this helps can i plz have brainlist :D hehe)

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Kwame rode bicycle for a distance of xkm and walked for another 1/2 hour at a rate of 6km/hour. If Kwame covered a total distanc

ElenaW [278]

1/2 hour at 6 km/ hour = 3 km.

Total distance was 10km, so he rode his bike for 10 - 3 = 7 km

x = 7

7 0

3 years ago

What is the value of the expression?<br><br> 5^2−3^0

lapo4ka [179]

Exponents come before subtraction. 5^2=25 and 3^0=1, so 25-1=24.

8 0

3 years ago

If a pair of regular dice are tossed once, use the expectation formula to determine the expected sum of the numbers on the upwar

Darya [45]

Answer:

The expected sum of the numbers on the upward faces of the two dice is 7.

Step-by-step explanation:

Consider the provided information.

If two pair of dice tossed the possible out comes are:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5,6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6,6)

Now we need to find the expected sum of the numbers on the upward faces of the two dice.

The expected sums can be:

Sum: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

Prob: 1/36, 2/36, 3/36, 4/36, 5/36, 6/36, 5/36, 4/36, 3/36, 2/36, 1/36

As we know that the expectation of experiment can be calculated as:

$P(S_1)\cdot S_1+P(S_2)\cdot S_2+........+P(S_n)\cdot S_n$

Here S represents the numerical outcomes and P(S) is the respective probability.

Substitute the respective values in the above formula.

$=2\times\frac{1}{36}+3\times\frac{2}{36}+4\times\frac{3}{36}+5\times\frac{4}{36}+6\times\frac{5}{36}+7\times\frac{6}{36}+8\times\frac{5}{36}+9\times\frac{4}{36}+10\times\frac{3}{36}+11\times\frac{2}{36}+12\times\frac{1}{36}\\=\frac{252}{36}\\=7$

Hence, the expected sum of the numbers on the upward faces of the two dice is 7.

8 0

4 years ago