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DiKsa [7]
3 years ago
7

Kim is sending postcards to her family. Each postcard is rectangular with an area of 12 square inches, and the length is 4 inche

s more than the width. Find the length and width of each postcard. If x represents the width of each postcard, select all of the equations that could be used to solve this problem. Answer Choices
Mathematics
1 answer:
Juli2301 [7.4K]3 years ago
8 0
12times four times the x with it
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D) A shopkeeper bought two radios for Rs 6,000. He sold one of them at a profit of 20℅
Orlov [11]

Answer:

2400 and 3600

Step-by-step explanation:

Radio 1 cost = x

Radio 2 cost= 6000- x

  • x*1.2+(6000-x)*0.9= 6000*1.02
  • 1.2x-0.9x+5400=6120
  • 0.3x= 6120-5400
  • 0.3x= 720
  • x=720/0.3
  • x= 2400
  • 6000- x= 3600
8 0
3 years ago
X² + 6x = 7<br> Can u please tell me the answer I need it fast
balandron [24]
The first x=1
the second x=-7
4 0
3 years ago
Read 2 more answers
A truck can be rented from Company A for $110 a day plus $0.50 per mile. Company B charges $80 a day plus $0.75 per mile to rent
bulgar [2K]

m=number of miles

A=$110+$0.50m

B=$80+$0.80m

$110+$0.50m=$80+$0.80m Subtract ($80+$0.50m)

$30=$0.30m Divide each side by $0.30

100=m ANSWER: More than 100 miles must be driven per day to make Company A a better deal.

3 0
3 years ago
The manager of a pizza chain in Albuquerque, New Mexico, wants to determine the average size of their advertised 16-inch pizzas.
erastova [34]

Answer:

n=50

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X=16.10 represent the sample mean for the sample  

\mu population mean (variable of interest)

s=1.8 represent the sample standard deviation

n=25 represent the sample size  

2) Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

We need to find the degrees of freedom given by:

df=n-1=25-1=24

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that z_{\alpha/2}=\pm 1.96

Since we assume that we are taking a bigger sample then we can replace the t distribution with the normal standard distribution, and we can assume that th population deviation is 1.8. The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

And on this case we have that ME =0.5 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

Replacing into formula (b) we got:

n=(\frac{1.96(1.8)}{0.5})^2 =49.79 \approx 50

So the answer for this case would be n=50 rounded up to the nearest integer

5 0
3 years ago
Simply the expressions by distributing and combining like terms , if necessary 16x + 5.6(2x - 11)
Anastasy [175]

Answer:

27.2x - 61.6

Step-by-step explanation:

16x + 5.6(2x - 11)

16x + 11.2x - 61.6

27.2x - 61.6

5 0
4 years ago
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