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3 years ago

C. 3 bananas cost $0.99. At this rate, how much do 5 bananas cost? Explain how you know.

Mathematics

2 answers:

djverab [1.8K]3 years ago

6 0

Answer:

$1.65

Step-by-step explanation:

You can get the cost of each banana by using the division operation. All you have to do is to divide $0.99 by 3.

0.99 ÷ 3 = $0.33

Each banana costs $0.33

Now that you know the cost of each banana, you may now get the cost of 5 bananas by using the multiplication operation. All you have to do is to multiply $0.33 by 5.

0.33 x 5 = $1.65

Therefore, 5 bananas cost $1.65.

Triss [41]3 years ago

3 0

The answer is $1.65

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Write a function whose graph passes through : (-7,0), (2,0), (4,0) The function is f(x)=______

strojnjashka [21]

Answer:

Step-by-step explanation:

f(x) = y = 0

so, f(x) = 0

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3 years ago

In the past, the average age of employees of a large corporation has been 40 years. Recently, the company has been hiring older

Viktor [21]

Answer:

$p_v =P(t_{(63)}>2.5)=0.0075$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the mean age is significantly higher than 45 years at 5% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X=45$ represent the mean height for the sample

$s=16$ represent the sample standard deviation for the sample

$n=64$ sample size

$\mu_o =40$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean age is higher than 40 years, the system of hypothesis would be:

Null hypothesis: $\mu \leq 40$

Alternative hypothesis: $\mu > 40$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{45-40}{\frac{16}{\sqrt{64}}}=2.5$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=64-1=63$

Since is a one right tailed test the p value would be:

$p_v =P(t_{(63)}>2.5)=0.0075$

Conclusion

6 0

3 years ago

20 people applied for a job. Everyone either has a school certificate or diploma or even both. If 14 have school certificates an

STatiana [176]

Given:

Either has a school certificate or diploma or even both = 20 people

Having school certificates = 14

Having diplomas = 11

To find:

The number of people who have a school certificate only.

Solution:

Let A be the set of people who have school certificates and B be the set of people who have diplomas.

According to the given information, we have

$n(A)=14$

$n(B)=11$

$n(A\cup B)=20$

We know that,

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$

$20=14+11-n(A\cap B)$

$20=25-n(A\cap B)$

Subtract both sides by 25.

$20-25=-n(A\cap B)$

$-5=-n(A\cap B)$

$5=n(A\cap B)$

We need to find the number of people who have a school certificate only, i.e. $n(A\cap B')$ .

$n(A\cap B')=n(A)-n(A\cap B)$

$n(A\cap B')=14-5$

$n(A\cap B')=9$

Therefore, 9 people have a school certificate only.

3 0

3 years ago

In a study of the impact of smoking on birth weight, researchers analyze birth weights (in grams) for babies born to 189 women w

kompoz [17]

Answer:

c. Smoking is associated with lower birth weights. When smokers are compared to nonsmokers, we are 95% confident that the mean weight of babies born to nonsmokers will be between 76.5 grams to 486.9 grams more than the mean birth weight of babies born to smokers.

Step-by-step explanation:

The difference in mean birth weights (nonsmokers minus smokers) is 281.7 grams with a margin of error of 205.2 grams with 95% confidence.

Then, we know that the 95% confidence interval is (76.5, 486.9)

$LL=M-MOE=281.7-205.2=76.5\\\\UL=M-MOE=281.7+205.2=486.9$

a. We are 95% confident that smoking causes lower birth weights by an average of between 76.5 grams to 486.9 grams.

False, it interprets the confidence interval as a probability for individual cases. The confidence interval is a range for the population mean.

b. There is a 95% chance that if a woman smokes during pregnancy her baby will weigh between 76.5 grams to 486.9 grams less than if she did not smoke.

False, it interprets the confidence interval as a probability for individual cases. The confidence interval is a range for the population mean.

c. Smoking is associated with lower birth weights. When smokers are compared to nonsmokers, we are 95% confident that the mean weight of babies born to nonsmokers will be between 76.5 grams to 486.9 grams more than the mean birth weight of babies born to smokers.

True.

d. With such a large margin of error, this study does not suggest that there is a difference in mean birth weights when we compare smokers to nonsmokers.

There is enough evidence, as the lower bound of the confidence is positive. This means that there is only a probability of 0.05/2=0.025 that the true mean weight difference is smaller than 76.5 grams.

6 0

3 years ago

For a class trip, the teachers would like to have one adult for every ten students. There are 190 students on the trip. How many

coldgirl [10]

You can solve this problem with the following

$\frac{1}{10} = \frac{a}{190}$
$10a = 190$
$a = 19$

There should be 19 adults on the trip.

6 0

3 years ago