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3 years ago

Maribel purchased 15 yards of fabric for a

Mathematics

2 answers:

steposvetlana [31]3 years ago

8 0

Answer:

Step-by-step explanation:

15-13=2

Phantasy [73]3 years ago

6 0

Answer:

2 yards

Step-by-step explanation:

subtract 13 from 15

15-13=2

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Which of the following lines is not parallel to the graph of y=4x+7. y-4x=9 4x-y=8 . 4y-x=8 8x-2y=8

DENIUS [597]

y = 4x + 7

This equation is in slope-intercept form, which is y = mx + b, where m is the slope and b is the y-intercept. The slope of this line is 4. Parallel lines have the same slope, thus a line not parallel to this line will not have a slope of 4. Let's check which of these lines does not have a slope of 4.

y - 4x = 9

y = 9 + 4x (add 4x to both sides)

y = 4x + 9 (switch the constant and x-term)

This line has a slope of 4.

4x - y = 8

-y = 8 - 4x (subtract 4x from both sides)

y = -8 + 4x (divide both sides by -1)

y = 4x - 8 (switch the constant and x-term)

This line has a slope of 4.

4y - x = 8

4y = 8 + x (add x to both sides)

y = 8/4 + x/4 (divide both sides by 4)

y = x/4 + 2 (simplify and switch constant and x-term)

This line has a slope of 1/4. It is thus not parallel to the line y = 4x + 7.

Answer:

4y - x = 8

5 0

3 years ago

For the years from 2002 and projected to 2024, the national health care expenditures H, in billions of dollars, can be modeled b

dmitriy555 [2]

Answer:

2019.

Step-by-step explanation:

We have been given that for the years from 2002 and projected to 2024, the national health care expenditures H, in billions of dollars, can be modeled by $H = 1,500e^{0.053t}$ where t is the number of years past 2000.

To find the year in which national health care expenditures expected to reach $4.0 trillion (that is, $4,000 billion), we will substitute $H=4,000$ in our given formula and solve for t as:

$4,000= 1,500e^{0.053t}$

$\frac{4,000}{1,500}=\frac{ 1,500e^{0.053t}}{1,500}$

$\frac{8}{3}=e^{0.053t}$

$e^{0.053t}=\frac{8}{3}$

Take natural log of both sides:

$\text{ln}(e^{0.053t})=\text{ln}(\frac{8}{3})$

$0.053t\cdot \text{ln}(e)=\text{ln}(\frac{8}{3})$

$0.053t\cdot (1)=0.9808292530117262$

$\frac{0.053t}{0.053}=\frac{0.9808292530117262}{0.053}$

$t=18.506212320$

So in the 18.5 years after 2000 the expenditure will reach 4 trillion.

$2000+18.5=2018.5$

Therefore, in year 2019 national health care expenditures are expected to reach $4.0 trillion.

7 0

3 years ago

Write a summary about decimals that are equivalent to fractions with denominators of 10 or 100. EXTRA CREDIT

Flura [38]

Decimals that contain a zero in the tenth position can be thought of as a fraction over 100 as the tenth isn't occupied and the hundredth correlates to the 100 as a denominator. The same goes for a number with the tenth position occupied. These decimals would be put over 10 as they are occupying that position in the decimal.

7 0

3 years ago

A 10-page play is $0.70 per page to e

Bas_tet [7]

Answer:

The answer of this question is 7

Step-by-step explanation:

because 1 page=$0.70 and now calculate

10×1×0.70=7

8 0

2 years ago

"The manager for State Bank and Trust has recently examined the credit card account balances for the customers of her bank and f

yuradex [85]

Answer:

$P(X = 4 ) = 0.1876$

$P(X \le 4) = 0.8358$

Step-by-step explanation:

From the question we are told that

The proportion that has outstanding balance is p = 0.20

The sample size is n = 15

Given that the properties of the binomial distribution apply, for a randomly selected number(X) of credit card

$X \ \ ~ Bin (n , p )$

Generally the probability of finding 4 customers in a sample of 15 who have "maxed out" their credit cards is mathematically represented as

$P(X = 4 ) = ^nC_4 * p^4 * (1 - p)^{n-4}$

=> $P(X = 4 ) = ^{15}C_4 * (0.20)^4 * (1 - 0.20)^{15-4}$

Here C stand for combination

=> $P(X = 4 ) = 0.1876$

Generally the probability that 4 or fewer customers in the sample will have balances at the limit of the credit card is mathematically represented as

$P(X \le 4) = [ ^{15}C_0 * (0.20)^0 * (1 - 0.20)^{15-0}]+[ ^{15}C_1 * (0.20)^1 * (1 - 0.20)^{15-1}]+\cdots+[ ^{15}C_4 * (0.20)^4 * (1 - 0.20)^{15-4}]$

=> $P(X \le 4) = 0.8358$

4 0

3 years ago