Graph the line through the points (-6,4) and (0,-4)

Lin's father is paying for a $20 meal. He has 20%-off coupon for the meal, After the discount, a 5% sales tax is applied. How mu

Serjik [45]

Answer:

$16.80

Step-by-step explanation:

20 % of 20 is 4 and you would take the 4 off for a discount. 5% of 16 is 0.8 and you would add it since it's an additional tax.

3 0

3 years ago

Four students have summer jobs. The table shows how much each student earns after working a certain number of hours. Which stude

mash [69]

Answer:

Step-by-step explanation:

7 0

3 years ago

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What is the value of x in the equation 2(x+3)=4(x-1)

marishachu [46]

Step 1: Distribute the 2 to numbers and variables inside parentheses

(2 × x) + (2 × 3) = 4 (x - 1)

2x + 6 = 4 (x - 1)

Step 2: Distribute the 4 to numbers and variables inside parentheses

2x + 6 = (4 × x) + ( 4 × - 1)

2x + 6 = 4x+ (-4)

2x + 6 = 4x- 4

Step 3: Combine like terms (x's go with x's) by subtracting 2x to both sides

(2x - 2x) + 6 = (4x - 2x) - 4

6 = 2x - 4

Step 4: Combine like terms by adding 4 to both sides

6 + 4 = 2x - 4 + 4

10 = 2x

Step 5: Isolate x by dividing 2 to both sides

10 ÷ 2 = 2x ÷2

5 = x

x = 5

To check: plug 5 into all the x's of the equation

2(x+3)=4(x-1)

2(5+3)=4(5-1)

2(8) = 4(4)

16 = 16

Hope this helped!

4 0

3 years ago

During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of str

Brilliant_brown [7]

Answer:

number of successes

$k = 235$

number of failure

$y = 265$

The criteria are met

A

The sample proportion is $\r p = 0.47$

B

$E =4.4 \%$

C

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from the true population proportion will not more than 4.4%

Ci

$r = 0.514 = 51.4 \%$

$v = 0.426 = 42.6 \%$

D

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

E

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

F

Yes our result would support the claim because

$\frac{1}{3 } \ of N < \frac{1}{2} (50\%) \ of \ N , \ Where\ N \ is \ the \ population\ size$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 500$

The sample proportion is $\r p = 0.47$

Generally the number of successes is mathematical represented as

$k = n * \r p$

substituting values

$k = 500 * 0.47$

$k = 235$

Generally the number of failure is mathematical represented as

$y = n * (1 -\r p )$

substituting values

$y = 500 * (1 - 0.47 )$

$y = 265$

for approximate normality for a confidence interval criteria to be satisfied

$np > 5 \ and \ n(1- p ) \ >5$

Given that the above is true for this survey then we can say that the criteria are met

Given that the confidence level is 95% then the level of confidence is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha =0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

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

Generally the margin of error is mathematically represented as

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

substituting values

$E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }$

$E = 0.044$

=> $E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p + E$

substituting values

$0.47 - 0.044 < p < 0.47 + 0.044$

$0.426 < p < 0.514$

The upper limit of the 95% confidence interval is $r = 0.514 = 51.4 \%$

The lower limit of the 95% confidence interval is $v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } < \frac{1}{2} (50\%)$

3 0

3 years ago

Which is a counterexample to this conjecture?

Svetradugi [14.3K]

The answer is C.

33,21,and 15 are all composite numbers. Meaning that they had factors other than 1 and themselves. 13 is a prime number because 13 only factors are 1 and itself.

3 0

3 years ago

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