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3 years ago

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While at the pumpkin farm, you also bought a taffy apple for $2.10, a pound of apples for $4.45 and a mum plant for $10.30. You

handed the cashier a $20. How much change should you receive?
A.$16.85
B.$3.15
C.$23.15
D.$36.85

2 answers:

lbvjy [14]3 years ago

5 0

Answer: B. $3.15

Step-by-step explanation: Hi! So first things you want to add all the things he bought together and then subtract it from the $20. I’m sorry if this isn’t right.

sleet_krkn [62]3 years ago

3 0

The answer is B
2.10+4.45+10.30=16.85 20-16.85=3.15
So B. $3.15
Hoped this helped:)

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An art history professor assigns letter grades on a test according to the following scheme. A: Top 13% of scores B: Scores below

madam [21]

Answer:

The numerical limits for a B grade are 81 and 89, that is, a score between 81 and 89 gets a B grade.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Scores on the test are normally distributed with a mean of 79.7 and a standard deviation of 8.4.

This means that $\mu = 79.7, \sigma = 8.4$

B: Scores below the top 13% and above the bottom 56%

So between the 56th percentile and the 100 - 13 = 87th percentile.

56th percentile:

X when Z has a p-value of 0.56, so X when Z = 0.15. Then

$Z = \frac{X - \mu}{\sigma}$

$0.15 = \frac{X - 79.7}{8.4}$

$X - 79.7 = 0.15*8.4$

$X = 81$

87th percentile:

X when Z has a p-value of 0.87, so X when Z = 1.13.

$Z = \frac{X - \mu}{\sigma}$

$1.13 = \frac{X - 79.7}{8.4}$

$X - 79.7 = 1.13*8.4$

$X = 89$

The numerical limits for a B grade are 81 and 89, that is, a score between 81 and 89 gets a B grade.

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2 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

three-quarters of students are wearing the school's colors. if there are 300 students how many are wearing the colors?

soldi70 [24.7K]

Answer:

225 students are wearing school colors.

Step-by-step explanation:

1. we know that 75% or .75 of the students are wearing school colors.

2. to find what this number is we need to know what 75% of 300 is.

("of" is a keyword meaning multiply)

3. We can do this by multiplying:

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3 years ago

A sample of 1200 computer chips revealed that 45% of the chips fail in the first 1000 hours of their use. The company's promotio

yaroslaw [1]

Answer:

$z=\frac{0.45 -0.48}{\sqrt{\frac{0.48(1-0.48)}{1200}}}=-2.08$

$p_v = P(Z$

So the p value obtained was a low value and using the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of chips that fail in the first 1000 hours of their use is not significantly less than 0.48.

Step-by-step explanation:

Data given and notation

n=1200 represent the random sample taken

$\hat p=0.45$ estimated proportion of chips that fail in the first 1000 hours of their use

$\mu_0 =0.48$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion si less then 0.48:

Null hypothesis: $p\geq 0.48$

Alternative hypothesis: $p < 0.48$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.45 -0.48}{\sqrt{\frac{0.48(1-0.48)}{1200}}}=-2.08$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v = P(Z$

So the p value obtained was a low value and using the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of chips that fail in the first 1000 hours of their use is not significantly less than 0.48.

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3 years ago