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3 years ago

If a sweatshirt is on sale for 35% and now costs $52, what was the original price?

2 answers:

8 0

The original price was $70.2. (If it’s not supposed to have cents then it’s just $70).

This is because you convert 35% to 1.35. Then, you multiply 52 by 1.35. That’ll give you 70.2.

CaHeK987 [17]3 years ago

8 0

Answer:

The original price of the sweatshirt is $80.

Step-by-step explanation:

If a product is on sale for a 35% discount, then it is 65% of the original price ( because 100 - 35 = 65)

So set up a proportion,

((part)/(whole)) = ((x)/(100))

In the given information, we know that the part (the amount paid) is 52$, what we need to find is the whole. To make solving the problem easier, let's call the whole x.

((52)/(x)) = ((65)/(100))

Simplify,

52/x = 13/20

Cross multiply,

1040 = 13x

Inverse operations,

1040 = 13x

/13 /13

80 = x

So the original price of the sweatshirt is $80.

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g A random sample of 100 students was taken. Eighty-five of the student in the sample experienced anxiety during the exam. We ar

BaLLatris [955]

Answer:

The test statistic is $z = 1.25$

Step-by-step explanation:

We are interested in determining whether or not the proportion of the student who experience anxiety during the exam is significantly more than 80%.

At the null hypothesis, we test if the proportion is 80%, that is:

$H_0: p = 0.8$

At the alternate hypothesis, we test if the proportion is more than 80%, that is:

$H_a: p > 0.8$

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.

80% is tested at the null hypothesis:

This means that $\mu = 0.8, \sigma = \sqrt{0.2*0.8} = 0.4$

A random sample of 100 students was taken. Eighty-five of the student in the sample experienced anxiety during the exam.

This means that $n = 100, X = \frac{85}{100} = 0.85$

The test statistic is

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

$z = \frac{0.85 - 0.8}{\frac{0.4}{\sqrt{10}}}$

$z = 1.25$

The test statistic is $z = 1.25$

4 0

3 years ago

Add x^3 - 4x^2 + 1 to 3x^2 + x Show Your work!

Sladkaya [172]

$\qquad \qquad\huge \underline{\boxed{\sf Answer}}$

Let's solve ~

$\qquad \sf \dashrightarrow \: (x {}^{3} - 4x {}^{2} + 1) + (3 {x}^{2} + x)$

$\qquad \sf \dashrightarrow \: {x}^{3} - 4 {x}^{2} + 1 + 3 {x}^{2} + x$

$\qquad \sf \dashrightarrow \: {x}^{3} - 4 {x}^{2} + 3x {}^{2} + x + 1$

$\qquad \sf \dashrightarrow \: {x}^{3} - {x}^{2} + x + 1$

6 0

2 years ago

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oksano4ka [1.4K]

Answer:

See below.

Step-by-step explanation:

1. 3 * 15 + 4 * 15 - 5 * 15

45 + 60 - 75

105 - 75

2. 3x + 4x - 5x

7x - 5x

3. 3(x-2) + 4(x-2) - 5(x-2)

3x - 6 + 4x - 8 - 5x + 10

2x - 4

4. (I'm not typing that out)

7.5x + 19.5 + 10x + 26 - 12.5x - 32.5

5x+13

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8 0

2 years ago

A_____________ is the numerical factor of a multiplication expression like 3x.

ch4aika [34]

The numerical factor of a multiplication expression like 3x is calied a coefﬁcient

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3 years ago

The rate at which rain accumulates in a bucket is modeled by the function r given by r(t)=10t−t^2, where r(t) is measured in mil

mars1129 [50]

Answer:

36 milliliters of rain.

Step-by-step explanation:

The rate at which rain accumluated in a bucket is given by the function:

$r(t)=10t-t^2$

Where r(t) is measured in milliliters per minute.

We want to find the total accumulation of rain from t = 0 to t = 3.

We can use the Net Change Theorem. So, we will integrate function r from t = 0 to t = 3:

$\displaystyle \int_0^3r(t)\, dt$

Substitute:

$=\displaystyle \int_0^3 10t-t^2\, dt$

Integrate:

$\displaystyle =5t^2-\frac{1}{3}t^3\Big|_0^3$

Evaluate:

$\displaystyle =(5(3)^2-\frac{1}{3}(3)^3)-(5(0)^2-\frac{1}{3}(0)^3)=36\text{ milliliters}$

36 milliliters of rain accumulated in the bucket from time t = 0 to t = 3.

4 0

3 years ago

If a sweatshirt is on sale for 35% and now costs $52, what was the original price?​

If a sweatshirt is on sale for 35% and now costs $52, what was the original price?