Which of the following uses logical deduction to prove the truth (or falsehood) of a specific statement?

What is 4 yards 1 foot and 3 inches minus 2 yards 1 foot and 9 inches?

alexdok [17]

1 yard 2 feet 6 inches

5 0

3 years ago

Write an equation of a line thru (3,-1) and (-1,-7)

Marrrta [24]

-1 - (-7) over 3 - (-1) = M

-1 + 7 over 3 + 1 = M

6 over 4 = M

6/4 = M

3/2 = M

6 0

4 years ago

Read 2 more answers

The chances of a tax return being audited are about 21 in 1 comma 000 if an income is less than $100,000 and 29 in 1 comma 000

vladimir2022 [97]

Answer:

a) p=0.021

b) p=0.029

c) Exactly one audit: P=0.0965

More than one audit: P=0.0042

Step-by-step explanation:

a) If the income is less than $100,000, there is a probability of 21 in 1,000 of being audited. This is:

$p_1=\dfrac{21}{1,000}=0.021$

If the income is equal or more than $100,000, there is a probability of 29 in 1,000 of being audited. This is:

$p_2=\dfrac{29}{1,000}=0.029$

b) If we have 5 taxpayers with incomes under $100,000, and we want to know the probability that exactly one will be audited, we can model this a binomial randome variable, with p=0.021 and n=5:

$P(x=1) = \dbinom{5}{1} p^{1}q^{4}=5*0.021*0.9186=0.0965\\\\$

There is a probability of 0.0965 that exactly one out of a sample of five taxpayers with incomes under $100,000 will be audited.

To calculate the probability that more than one will be audited, we use the same distribution:

$P(x>1)=1-[P(x=0)+P(x=1)]\\\\\\P(x=0) = \dbinom{5}{0} p^{0}q^{5}=1*1*0.8993=0.8993\\\\\\P(x=1) = \dbinom{5}{1} p^{1}q^{4}=5*0.021*0.9186=0.0965\\\\\\P(x>1)=1-[P(x=0)+P(x=1)]\\\\P(x>1)=1-(0.8993+0.0965)\\\\P(x>1)=1-0.9958\\\\P(x>1)=0.0042$

There is a probability of 0.0042 that more than one out of a sample of five taxpayers with incomes under $100,000 will be audited.

8 0

3 years ago

Christopher wrote the number pattern below. The first term is eight. 8, six, nine, seven, ten is the pattern what is the rule fo

Nataly [62]

-2 then +3 then repeat

7 0

3 years ago

Read 2 more answers

In recent years, only 79% of the fruit sold at Fred's Fruit stand has been any good. Fred recently started buying fruit from new

Juli2301 [7.4K]

Answer:

The p-value of the test is 0.0485 < 0.05, also less than 0.07, so there is sufficient evidence to conclude that there is a statistically significant increase in the percentage of good fruit, for both significance levels, thus the change in significance levels do not change the conclusion.

Step-by-step explanation:

79% of the fruit sold at Fred's Fruit stand has been any good. Test if there has been an increase.

At the null hypothesis, we test if there has been no increase, that is, the proportion is still of 79%, so:

$H_0: p = 0.79$

At the alternative hypothesis, we test if there has been an increase, that is, more than 79% being good, so:

$H_1: p > 0.79$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.79 is tested at the null hypothesis:

This means that $\mu = 0.79, \sigma = \sqrt{0.79*0.21}$

In a random sample if 475 pieces, 85 were bad.

So 475 - 85 = 390 were good, and:

$n = 475, X = \frac{390}{475} = 0.8211$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.8211 - 0.79}{\frac{\sqrt{0.79*0.21}}{\sqrt{475}}}$

$z = 1.66$

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion above 0.8211, which is 1 subtracted by the p-value of Z = 1.66.

Looking at the z-table, z = 1.66 has a p-value of 0.9515.

1 - 0.9515 = 0.0485

The p-value of the test is 0.0485 < 0.05, also less than 0.07, so there is sufficient evidence to conclude that there is a statistically significant increase in the percentage of good fruit, for both significance levels, thus the change in significance levels do not change the conclusion.

5 0

3 years ago