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3 years ago

14

For the equation, x-7=24, can a value be substituted for x to make that a true number sentence? How many values could be substit

uted for x and have a true number sentence?

2 answers:

ra1l [238]3 years ago

6 0

We add 7 to both sides, making x=31. we can substitute 31 and make the true number sentence.
only 31 can be substituted, as it is only one x value, holding only one true value, which we found out is 31.
hope this helped!! xx

chubhunter [2.5K]3 years ago

6 0

Answer:

1 variable or 31

Step-by-step explanation:

7+24=31 giving you the only variable

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Fewer young people are driving. In , of people under years old who were eligible had a driver's license. Bloomberg reported that

insens350 [35]

Complete Question

Fewer young people are driving. In 1995, 63.9% of people under years 20 old who were eligible had a driver's license. Bloomberg reported that percentage had dropped to 41.7% in 2016. Suppose these results are based on a random sample 1,200 of people under 20 years old who were eligible to have a driver's license in 1995 and again in 2016.

a. At 95% confidence, what is the margin of error and the interval estimate of the number of eligible people under 20 years old who had a driver's license in 1995?

Margin of error(to four decimal places)

Interval estimate (to four decimal places)

b. At 95% confidence, what is the margin of error and the interval estimate of the number of eligible people under 20 years old who had a driver's license in 2016?

Margin of error(to four decimal places)

Interval estimate to (to four decimal places)

Answer:

a

$0.6120 < p < 0.639 + 0.6670$

b

$0.3900 < p < 0.4440$

Step-by-step explanation:

Considering question a

The sample proportion is 1995 is $\^ p_1 = 0.639$

The sample size is $n = 1200$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }$

=> $E = 1.96 * \sqrt{\frac{0.639 (1- 0.639)}{1200} }$

=> $E = 0.027$

Generally 95% Interval estimate is mathematically represented as

$\^ p -E < p < \^ p +E$

=> $0.639 -0.027 < p < 0.639 + 0.027$

=> $0.6120 < p < 0.639 + 0.6670$

Considering question b

The sample proportion is 1995 is $\^ p_2 = 0.417$

The sample size is $n = 1200$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }$

=> $E = 1.96 * \sqrt{\frac{0.417 (1- 0.417)}{1200} }$

=> $E = 0.027$

Generally 95% Interval estimate is mathematically represented as

$\^ p -E < p < \^ p +E$

=> $0.417 -0.027 < p < 0.417 + 0.027$

=> $0.3900 < p < 0.4440$

6 0

3 years ago

Jeff is baking a cake. The recipe says that he has to mix 32 grams of vanilla powder to the flour. Jeff knows that 1 cup of that

ValentinkaMS [17]

1 / 128 = (2/3) / x....1 cup to 128 grams = 2/3 cups to x grams
cross multiply
(1)(x) = (128)(2/3)
x = 256/3
x = 85.3

he went over...
85.3 - 32 = 53.3....he went over by 53.3 (or 53 1/3) grams

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4 years ago

What is the domain of the relation?

Thepotemich [5.8K]

Answer: $-4$

Step-by-step explanation:

Domain: x-axis

Range: y-axis

The relation graph shows the points in which the domain is located. To find the Domain, go to the point in the x-axis (the horizontal line) and note the number where the point lies. For this question, the point on the left side of the graph lies on negative four ( $-4$ ), and on the right side, the point is on 6. Therefore, the domain of this relation is negative four is greater than/equal to x is less than/equal to six. It could also be written like this: $-4\leq x\leq 6$ .

Learn more: brainly.com/question/24574301

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3 years ago

Sali throws an ordinary fair 6 sided dice once.

xenn [34]

a) 1/6

b) 1/36

c)

1H, 2H, 3H, 4H, 5H, 6H

1T, 2T, 3T, 4T, 5T, 6T

Step-by-step explanation:

a)

The probability of a certain event A to occur is given by

$p(A)=\frac{a}{n}$

where

a is the number of successfull outcomes, in which event A occurs

n is the total number of possible outcomes

In this problem, the event is

"getting a 6 when throwing a dice once"

We know that the possible outcomes of a dice are six: 1, 2, 3, 4, 5, 6, so we he have

$n=6$

The successfull outcome in this case is only if we get a 6, so only 1 outcome, therefore

$a=1$

So, the probability of this event is

$p(6)=\frac{1}{6}$

b)

In this case instead, we are throwing the dice twice.

The two throws of the dice are independent events (one does not depend on the other): the probability that two independent events A and B occur at the same time is given by the product of the individual probabilities,

$p(AB)=p(A)\cdot p(B)$

where

$p(A)$ is the probability that event A occurs

$p(B)$ is the probability that event B occurs

Here we have:

- Event A is "getting a 6 in the first throw of the dice". We already calculated this probability in part a), and it is

$p(A)=\frac{1}{6}$

- Event B is "getting a 6 in the second throw of the dice". Since the dice has not changed, the probability is still the same, so

$p(B)=\frac{1}{6}$

Therefore, the probability of getting a 6 on both throws is:

$p(66)=p(6)\cdot p(6)=\frac{1}{6}\cdot \frac{1}{6}=\frac{1}{36}$

c)

In this problem, we have:

- A dice that is thrown once

- A coin that is also thrown once

The dice has 6 possible outcomes, as we stated in part a):

1, 2, 3, 4, 5, 6

While the coin has two possible outcomes:

H = head

T = tail

So, in order to find all the outcomes of the two events combined, we have to combine all the outcomes of the dice with all the outcomes of the coin.

Doing so, using the following notation:

1H (getting 1 with the dice, and head with the coin)

The possible outcomes are:

1H, 2H, 3H, 4H, 5H, 6H

1T, 2T, 3T, 4T, 5T, 6T

So, we have a total of 12 possible outcomes.

4 0

3 years ago

I need help with my math class

aivan3 [116]

Answer:

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3 years ago

Read 2 more answers