(i’ll mark brainliest if it’s correct )

Determine the Type I error if the null hypothesis, H0, is: an electrician claims that no more than 10% of homes in the city are

kap26 [50]

Answer : There are no more than 10% of homes in the city are not up to the current electric codes but actually we do not have evidence to accept it.

Explanation:

The Type I error occurs when the null hypothesis is true but we reject it.

Given null hypothesis : $H_0$ : an electrician claims that no more than 10% of homes in the city are not up to the current electric codes.

The Type I error for the given situation is that there are no more than 10% of homes in the city are not up to the current electric codes but actually we do not have evidence to accept it.

3 0

3 years ago

Subtract.

mafiozo [28]

D) 3x+6

x + 1 + 2x +5

That will = D

4 0

3 years ago

Read 2 more answers

Consider a game in which players roll a number cube to determine the number of points earned. If a player rolls a prime number,

Aleksandr-060686 [28]

Answer:

The expected value of the points earned on a single roll in this game is $\dfrac{1}{6} = 0.1667$ .

Step-by-step explanation:

We are given that consider a game in which players roll a number cube to determine the number of points earned. If a player rolls a prime number, that many points will be added to the player’s total. Any other roll will be deducted from the player’s total.

Assuming that the numbered cube is a dice with numbers (1, 2, 3, 4, 5, and 6).

Here, the prime numbers are = 1, 2, 3 and 5

Numbers which are not prime = 4 and 6

This means that if the dice got the number 1, 2, 3 or 5, then that many points will be added to the player’s total and if the dice got the number 4 or 6, then that many points will get deducted from the player’s total.

Here, we have to make a probability distribution to find the expected value of the points earned on a single roll in this game.

Note that the probability of getting any of the specific number on the dice is $\dfrac{1}{6}$ .

Numbers on the dice (X) P(X)

+1 $\frac{1}{6}$

+2 $\frac{1}{6}$

+3 $\frac{1}{6}$

-4 $\frac{1}{6}$

+5 $\frac{1}{6}$

-6 $\frac{1}{6}$

Here (+) sign represent the addition in the player's total and (-) sign represents the deduction in the player's total.

Now, the expected value of X, E(X) = $\sum X \times P(X)$

= $(+1) \times \frac{1}{6} +(+2) \times \frac{1}{6} +(+3) \times \frac{1}{6} +(-4) \times \frac{1}{6} +(+5) \times \frac{1}{6} +(-6) \times \frac{1}{6}$