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3 years ago

A newspaper stand sold 1000 copies of the city newspaper for $1600. Write and solve an equation to find the price of one copy.

Mathematics

2 answers:

Goshia [24]3 years ago

8 0

Simple. just do this:
1000x=1600
x=1600÷1000
x=1.6

each newspaper is $1.60

svet-max [94.6K]3 years ago

7 0

Answer:

The correct answer is $1,60.

Step-by-step explanation:

Analyzing the information we have we can say that:

1000 copies of the city newspaper cost $ 1600.

Now we must find out how much each copy cost.

A good way to find out this would be through a cross-multiplication.

Where x will be the value we have to find out:

if 1000 copies ----- 1600

1 copy ---------- x

1000 x = 1600. 1

x = 1600: 1000

x = 1.60

This way we can show that a copy cost $ 1.60.

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A sum amount to rupees 34476 in two and a half years at 4% CI. the sum is

Kruka [31]

Given:

Amount = Rs. 34476

Rate of compound interest = 4%

Time = $2\dfrac{1}{2}=2.5$ years

To find:

The principal value.

Solution:

Formula for amount is

$A=P\left(1+\dfrac{r}{100}\right)^t$

Where, P is principal value, r is rate of interest and t is time in years.

Putting the given values, we get

$34476=P\left(1+\dfrac{4}{100}\right)^{2.5}$

$34476=P\left(1+0.04\right)^{2.5}$

$34476=P\left(1.04\right)^{2.5}$

$\dfrac{34476}{\left(1.04\right)^{2.5}}=P$

Now,

$P=31256.0090$

$P\approx 31256$

Therefore, the value of sum or principal value is Rs.31256.

6 0

2 years ago

The mean number of words per minute (WPM) read by sixth graders is 8888 with a standard deviation of 1414 WPM. If 137137 sixth g

Bingel [31]

Noticing that there is a pattern of repetition in the question (the numbers are repeated twice), we are assuming that the mean number of words per minute is 88, the standard deviation is of 14 WPM, as well as the number of sixth graders is 137, and that there is a need to estimate the probability that the sample mean would be greater than 89.87.

Answer:

"The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

Step-by-step explanation:

This is a problem of the distribution of sample means. Roughly speaking, we have the probability distribution of samples obtained from the same population. Each sample mean is an estimation of the population mean, and we know that this distribution behaves normally for samples sizes equal or greater than 30 $\\ n \geq 30$ . Mathematically

$\\ \overline{X} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$ [1]

In words, the latter distribution has a mean that equals the population mean, and a standard deviation that also equals the population standard deviation divided by the square root of the sample size.

Moreover, we know that the variable Z follows a normal standard distribution, i.e., a normal distribution that has a population mean $\\ \mu = 0$ and a population standard deviation $\\ \sigma = 1$ .

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [2]

From the question, we know that

The population mean is $\\ \mu = 88$ WPM
The population standard deviation is $\\ \sigma = 14$ WPM

We also know the size of the sample for this case: $\\ n = 137$ sixth graders.

We need to estimate the probability that a sample mean being greater than $\\ \overline{X} = 89.87$ WPM in the distribution of sample means. We can use the formula [2] to find this question.

The probability that the sample mean would be greater than 89.87 WPM

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

$\\ Z = \frac{89.87 - 88}{\frac{14}{\sqrt{137}}}$

$\\ Z = \frac{1.87}{\frac{14}{\sqrt{137}}}$

$\\ Z = 1.5634 \approx 1.56$

This is a standardized value and it tells us that the sample with mean 89.87 is 1.56 standard deviations above the mean of the sampling distribution.

We can consult the probability of P(z<1.56) in any cumulative standard normal table available in Statistics books or on the Internet. Of course, this probability is the same that $\\ P(\overline{X} < 89.87)$ . Then

$\\ P(z$

However, we are looking for P(z>1.56), which is the complement probability of the previous probability. Therefore

$\\ P(z>1.56) = 1 - P(z$

$\\ P(z>1.56) = P(\overline{X}>89.87) = 0.0594$

Thus, "The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

5 0

3 years ago

The data show systolic and diastolic blood pressure of certain people. Find the regression equation, letting the first variable

MakcuM [25]

Full Question:

The data show systolic and diastolic blood pressure of certain people. Find the regression equation, letting the first variable be the independent (x) variable. Find the best predicted diastolic pressure for a person with a systolic reading of 113. use a significance level of 0.05.

Systolic| 150 129 142 112 134 122 126 120

Diastolic| 88 96 106 80 98 63 95 64

a. What is the regression equation?

^y = __ + __x (Round to two decimal places as needed.)

b. What is the best predicted value?

^y is about __ (Round to one decimal place as needed.)

Answer:

A. yhat = a + bx = -10.64 + 0.75x

B. 74.0

Step-by-step explanation:

A. To find the regression equation here, we apply the formulas and then apply it to find the value of y given value of x:

calculate xbar and ybar which is the average of the variables:

Where n(number of values in x or y)=8

xbar = sum of x/n = 129.375

ybar = sum of y/n = 86.25

to calculate b

b= [Sum x^2 * Sum y - Sum x * Sum x*y] / [N*Sum x^2 - (Sum x)^2]

b = 0.74891

To calculate a

a = ybar - b * xbar = -10.64023

Regression equation:

y=mx+b= -10.64 + 0.75x

B. given x = 113,

y = -10.64023 + 0.74891 * 113

y= 74.0

5 0

3 years ago

The perimeter of this rectangle is 24cm. If the length is 8 cm, what is the height?

Gemiola [76]

Answer:

Step-by-step explanation:

P = s+s+s+s

length = 8

8x2=16

24-16=8

8 / 2 = 4

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3 years ago

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Sedaia [141]

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3 years ago

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