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Scrat [10]
3 years ago
8

A. 18 is 20 percent of what number? B. 15 is 25 percent of what number? C. 12 is 30 percent of what number? D. 24 is 40 percent

of what number? E. 39 is 60 percent of what number? F. 78 is 30 percent of what number? Question: Represent each as an equation with x as the unknown number, and solve.
Mathematics
1 answer:
Kisachek [45]3 years ago
8 0

A. 18 is 20% of 90

B. 15 is 25% of 60

C. 12 is 30% of 40

D.  24 is 40% of 60

E. 39 is 60% of 65

F. 78 is 30% of 234

(Sorry I have to go, but hopefully I helped you)

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Marisol buys 3 notebooks that each cost the same amount and 1 calculator that costs $9. The total cost is $21. How much does eac
adoni [48]

Answer: 4 dollars for each notebook

Step-by-step explanation:

21-9= 12

12/3=4

each notebooks costs 4 dollars

4 0
3 years ago
Read 2 more answers
Connecticut families were asked how much they spent weekly on groceries. Using the following data, construct and interpret a 95%
Amanda [17]

Answer:

The 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

Step-by-step explanation:

The data for the amount of money spent weekly on groceries is as follows:

S = {210, 23, 350, 112, 27, 175, 275, 50, 95, 450}

<em>n</em> = 10

Compute the sample mean and sample standard deviation:

\bar x =\frac{1}{n}\cdot\sum X=\frac{ 1767 }{ 10 }= 176.7

s= \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} }       = \sqrt{ \frac{ 188448.1 }{ 10 - 1} } \approx 144.702

It is assumed that the data come from a normal distribution.

Since the population standard deviation is not known, use a <em>t</em> confidence interval.

The critical value of <em>t</em> for 95% confidence level and degrees of freedom = n - 1 = 10 - 1 = 9 is:

t_{\alpha/2, (n-1)}=t_{0.05/2, (10-1)}=t_{0.025, 9}=2.262

*Use a <em>t</em>-table.

Compute the 95% confidence interval for the population mean amount spent on groceries by Connecticut families as follows:

CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\ \frac{s}{\sqrt{n}}

     =176.7\pm 2.262\cdot\ \frac{144.702}{\sqrt{10}}\\\\=176.7\pm 103.5064\\\\=(73.1936, 280.2064)\\\\\approx (73.20, 280.21)

Thus, the 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

7 0
3 years ago
What is the effect on the graph of the function f(x) = 2x when f(x) is replaced with f(0.5x)? A) vertical stretching B) vertical
PtichkaEL [24]
<h3><u>Answer:</u></h3>

Option: D

Horizontal stretching.

<h3><u>Step-by-step explanation:</u></h3>

We have to find the effect on the graph of the function f(x)=2x when it is replaced by f(0.5 x).

We know that when a parent function f(x) is replaced by f(kx) then either the graph is stretched horizontally or shrinked horizontally.

if k>1 then the graph is shrinked horizontally.

if k<1 then the graph is stretched horizontally.

Hence here k=0.5<1 so the graph of the function is stretched horizontally.

3 0
3 years ago
Read 2 more answers
A professor pays 25 cents for each blackboard error made in lecture to the student who pointsout the error. In a career ofnyears
marta [7]

Answer:

(a) The probability that <em>Y</em>₂₀ exceeds 1000  is 3.91 × 10⁻⁶.

(b) <em>n</em> = 28.09

Step-by-step explanation:

The random variable <em>Y</em>ₙ is defined as the total numbers of dollars paid in <em>n</em> years.

It is provided that <em>Y</em>ₙ can be approximated by a Gaussian distribution, also known as Normal distribution.

The mean and standard deviation of <em>Y</em>ₙ are:

\mu_{Y_{n}}=40n\\\sigma_{Y_{n}}=\sqrt{100n}

(a)

For <em>n</em> = 20 the mean and standard deviation of <em>Y</em>₂₀ are:

\mu_{Y_{n}}=40n=40\times20=800\\\sigma_{Y_{n}}=\sqrt{100n}=\sqrt{100\times20}=44.72\\

Compute the probability that <em>Y</em>₂₀ exceeds 1000 as follows:

P(Y_{n}>1000)=P(\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}>\frac{1000-800}{44.72})\\=P(Z>  4.47)\\=1-P(Z

**Use a <em>z </em>table for probability.

Thus, the probability that <em>Y</em>₂₀ exceeds 1000  is 3.91 × 10⁻⁶.

(b)

It is provided that P (<em>Y</em>ₙ > 1000) > 0.99.

P(Y_{n}>1000)=0.99\\1-P(Y_{n}

The value of <em>z</em> for which P (Z < z) = 0.01 is 2.33.

Compute the value of <em>n</em> as follows:

z=\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}\\2.33=\frac{1000-40n}{\sqrt{100n}}\\2.33=\frac{100}{\sqrt{n}}-4\sqrt{n}  \\2.33=\frac{100-4n}{\sqrt{n}} \\5.4289=\frac{(100-4n)^{2}}{n}\\5.4289=\frac{10000+16n^{2}-800n}{n}\\5.4289n=10000+16n^{2}-800n\\16n^{2}-805.4289n+10000=0

The last equation is a quadratic equation.

The roots of a quadratic equation are:

n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

a = 16

b = -805.4289

c = 10000

On solving the last equation the value of <em>n</em> = 28.09.

8 0
3 years ago
Mr. Francis bought a pair of shoes on sale for 40% off. The regular price was $55, not including tax. How much did Mr. Francis p
maw [93]

Answer:

He paid $33 dollars for the shoes

Step-by-step explanation:

8 0
3 years ago
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