I need help on these three PLEASE!!

Wilma buys a new game that is priced $43.89. She gets successive discounts of 15% followed by 5% off on the game. The purchase p

jek_recluse [69]

The answer is C. 80.75%

Hope this helps!

4 0

3 years ago

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On a graphing calculator, you can use the function normalcdf(lower bound, upper bound, μ, σ) to find the area under a normal cur

larisa86 [58]

Answer:

99.87% of cans have less than 362 grams of lemonade mix

Step-by-step explanation:

Let the the random variable X denote the amounts of lemonade mix in cans of lemonade mix . The X is normally distributed with a mean of 350 and a standard deviation of 4. We are required to determine the percent of cans that have less than 362 grams of lemonade mix;

We first determine the probability that the amounts of lemonade mix in a can is less than 362 grams;

Pr(X<362)

We calculate the z-score by standardizing the random variable X;

Pr(X<362) = $Pr(Z$

This probability is equivalent to the area to the left of 3 in a standard normal curve. From the standard normal tables;

Pr(Z<3) = 0.9987

Therefore, 99.87% of cans have less than 362 grams of lemonade mix

4 0

3 years ago

Tain has a great literary tradition that spans centuries. One might assume, then, that Britons read more than citizens of other

tresset_1 [31]

Answer:

Null hypothesis: $p_{1} \leq p_{2}$

Alternative hypothesis: $p_{1} > p_{2}$

$z=3.02$

$p_v =P(Z>3.02)=0.00127$

The p value is a very low value and using any significance level for example $\alpha=0.05, 0,1,0.15$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of Canadians is not significantly higher than the porportions of readers at Britons.

Step-by-step explanation:

1) Data given and notation

$X_{1}=0.86*1004$ represent the number of Canadians randomly sampled by Gallup that read at least one book in the past year

$X_{2}=0.81*1009$ represent the number of Britons randomly sampled that read at least one book in the past year

$n_{1}=1004$ sample of Gallup selected

$n_{2}=1009$ sample of Britons selected

$p_{1}=0.86$ represent the proportion of Canadians randomly sampled by Gallup that read at least one book in the past year

$p_{2}=0.81$ represent the proportion of Britons randomly sampled that read at least one book in the past year

z would represent the statistic (variable of interest)

$p_v$ represent the value for the test (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the proportion for men with red/green color blindness is a higher than the rate for women , the system of hypothesis would be:

Null hypothesis: $p_{1} \leq p_{2}$

Alternative hypothesis: $p_{1} > p_{2}$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{0.81+0.86}{2}=0.835$

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.86-0.81}{\sqrt{0.835(1-0.835)(\frac{1}{1004}+\frac{1}{1009})}}=3.02$

4) Statistical decision

For this case we don't have a significance level provided $\alpha$ , but we can calculate the p value for this test.

Since is a one side test the p value would be:

$p_v =P(Z>3.02)=0.00127$

So the p value is a very low value and using any significance level for example $\alpha=0.05, 0,1,0.15$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of Canadians is not significantly higher than the porportions of readers at Britons.

5 0

3 years ago

Hellohello, if you able to help me then please do. (:

timurjin [86]

Answer:

g=12

Step-by-step explanation:

3 + g = 15

g = 15-3

g = 12

5 0

2 years ago