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Pani-rosa [81]
3 years ago
10

A spinner is divided into three equal sections labeled a B and C. Erik spins the spinner and rolls a fair number cube. The tree

diagram shows the sample space for the possible outcomes. Which of the following statements is correct?
The probability of the spinner landing on B or rolling a 4 is 1/8
The probability of the spinner not landing on C and rolling a 3 is 13/18
The probability of the spinner landing on an A or rolling a number greater than two is 2/9
The probability of the spinner not landing on A and rolling a prime number is 1/3
Mathematics
2 answers:
ale4655 [162]3 years ago
7 0

Answer:

Last one:

The probability of the spinner not landing on A and rolling a prime number is 1/3

Step-by-step explanation:

The probability of the spinner not landing on A: 2/3

Prime outcomes on a dice:

2, 3, 5

The probability of rolling a prime number: 3/6 = 1/2

The probability of the spinner not landing on A and rolling a prime number is:

2/3 × 1/2

1/3

skelet666 [1.2K]3 years ago
3 0

Answer:

Last one:

The probability of the spinner not landing on A and rolling a prime number is 1/3

Step-by-step explanation:

The probability of the spinner not landing on A: 2/3

Prime outcomes on a dice:

2, 3, 5

The probability of rolling a prime number: 3/6 = 1/2

The probability of the spinner not landing on A and rolling a prime number is:

2/3 × 1/2

1/3

Step-by-step explanation:

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Exercise 11.21) Many smartphones, especially those of the LTE-enabled persuasion, have earned a bad rap for exceptionally bad ba
stiv31 [10]

Answer:

Step-by-step explanation:

Hello!

The researcher suspects that the battery life between charges for the Motorola Droid Razr Max differs if its primary use is talking or if its primary use is for internet applications.

Since the means for talk time usage (20hs) is greater than the mean for internet usage (7hs) the main question is if the variance in hours of usage is also greater when the primary use is talk time.

Be:

X₁: Battery duration between charges when the primary usage of the phone is talking. (hs)

n₁= 12

X[bar]₁= 20.50 hs

S₁²= 199.76hs² (S₁= 14.13hs)

X₂: Battery duration between charges when the primary usage of the phone is internet applications.

n₂= 10

X[bar]₂= 8.50

S₂²= 33.29hs² (S₂= 5.77hs)

Assuming that both variables have a normal distribution X₁~N(μ₁;σ₁²) and X₂~N(μ₂; σ₂²)

The parameters of interest are σ₁² and σ₂²

a) They want to test if the population variance of the duration time of the battery when the primary usage is for talking is greater than the population variance of the duration time of the battery when the primary usage is for internet applications. Symbolically: σ₁² > σ₂² or since the test to do is a variance ratio: σ₁²/σ₂² > 1

The hypotheses are:

H₀: σ₁²/σ₂² ≤ 1

H₁: σ₁²/σ₂² > 1

There is no level of significance listed so I've chosen α: 0.05

b) I've already calculated the sample standard deviations using a software, just in case I'll show you how to calculate them by hand:

S²= \frac{1}{n-1}*[∑X²-(∑X)²/n]

For the first sample:

n₁= 12; ∑X₁= 246; ∑X₁²= 7240.36

S₁²= \frac{1}{11}*[7240.36-(246)²/12]= 199.76hs²

S₁=√S₁²=√199.76= 14.1336 ≅ 14.13hs

For the second sample:

n₂= 10; ∑X₂= 85; ∑X₂²= 1022.12

S₂²= \frac{1}{9}*[1022.12-(85)²/10]= 33.2911hs²

S₂=√S₂²=√33.2911= 5.7698 ≅ 5.77hs

c)

For this hypothesis test, the statistic to use is a Snedecors F:

F= \frac{S_1^2}{S_2^2} *\frac{Sigma_1^2}{Sigma_2^2} ~~F_{n_1-1;n_2-1}

This test is one-tailed right, wich means that you'll reject the null hypothesis to big values of F:

F_{n_1-1;n_2-1; 1-\alpha }= F_{11;9;0.95}= 3.10

The rejection region is then F ≥ 3.10

F_{H_0}= \frac{199.76}{33.29} * 1= 6.0006

p-value: 0.006

Considering that the p-value is less than the level of significance, the decision is to reject the null hypothesis.

Then at a 5% level, there is significant evidence to conclude that the population variance of the duration time of the batteries of Motorola Droid Razr Max smartphones used primary for talk is greater than the population variance of the duration time of the batteries of Motorola Droid Razr Max smartphones used primary for internet applications.

I hope this helps!

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Answer:

The sum is 10(16 + 54) =700

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

It is always going to be 0 because there is always going to be a 2 and a 5 that multiply to make the units digit zero.

I hope this could help you.

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