Mona wrote 7 tests. her average was 93 what is the lowest grade she can get for her average to be 90

In this problem the number we are working with is:

105,159

By definition we note:

thousand place: a five-digit quantity greater than zero.

Moreover, the rounding rule is:

if the digit being removed is 5 or more, increase the kept digit by one.

Therefore, rounding to the nearest ten thousand yields:

105,159 = 110,000

Answer:

105,159 rounded to the nearest ten thousand is:

105,159 = 110,000

Answer:

E. P(W | H)

Step-by-step explanation:

Meaning of each probability:

P(H): Likelihood of the game occurring at home

P(W): Likelihood of the game resulting in a win.

P(H and W): Likelihood that the game is both at home and a win.

P(H|W): Likelihood of a win occurring when at home.

P(W|H): Likelihood of winning at home games.

Which probabilities are necessary to determine the fraction of home games that were victories?

This is the probability of winning a home game. Hence, the answer is:

E. P(W | H)

Approximate length of wire ≈ 193 ft (to the nearest foot).

Answer:

a) 2.60x10^-7

b) 5.31x10^-5

c) 2.19x10^-3

Step-by-step explanation:

X=the number of hits

Probability can be determined by taking the number of successful outcomes and dividing it by the total number of possible outcomes.

Now

a) P(X=6)=P({1, 1, 1, 1, 1, 1})=6/40*5/39*4/38*3/37*2/36*1/35=2.60x10^-7

b) P(X=5)=P({0, 1, 1, 1, 1, 1})+P({1, 0, 1, 1, 1, 1})+...+P({1, 1, 1, 1, 1, 0}), and each scenario carries the same probability

P(X=5)=6P({1, 1, 1, 1, 1, 0})=6*(6/40*5/39*4/38*3/37*2/36*34/35)=5.31x10^-5

c) P(X=4)=P({0, 0, 1, 1, 1, 1})+...+P({1, 1, 1, 1, 0, 0}), where each case has an identical probability.

I am not very familiar with this topic, but it seems to me that the average is around 8, which would suggest a deviation of approximately 2.