The rectangle is 3 3⁄4 centimeters long and 2 1⁄2 centimeters wide. What is the area of this rectangle?

Is this correct ? if not what’s the correct answer?

lara [203]

Yes this is correct because that angle in is with the other angles to

5 0

2 years ago

Read 2 more answers

Please help with question got stuck

julsineya [31]

Answer:

y=4x

Step-by-step explanation:

8 0

3 years ago

In 1985 there were 275 cell phone subscribers in the small town of China grove. Subscribers increased by 40% each year after 198

emmasim [6.3K]

1985year ->275cell phones->100%
1994year -> x cell phones ->100%+360%
40% • (1994-1985)= 40% • 9= 360%

275 cell phone 100%
X cell phones. 100+360%=460%
x=275•460%/100%
x=12420/100
x=1265
99% sure this is a correct answer

x=275 • 460%/100%
x=12,420/100
x=1265

1994-1985= 9 yrs • 40%= 360%

5 0

3 years ago

The distribution of lifetimes of a particular brand of car tires has a mean of 51,200 miles and a standard deviation of 8,200 mi

Orlov [11]

Answer:

a) 0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

b) 0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

c) 0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

d) 0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

Step-by-step explanation:

Problems of normally distributed distributions are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 51200, \sigma = 8200$

Probabilities:

A) Between 55,000 and 65,000 miles

This is the pvalue of Z when X = 65000 subtracted by the pvalue of Z when X = 55000. So

X = 65000

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{65000 - 51200}{8200}$

$Z = 1.68$

$Z = 1.68$ has a pvalue of 0.954

X = 55000

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{55000 - 51200}{8200}$

$Z = 0.46$

$Z = 0.46$ has a pvalue of 0.677

0.954 - 0.677 = 0.277

0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

B) Less than 48,000 miles

This is the pvalue of Z when X = 48000. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{48000 - 51200}{8200}$

$Z = -0.39$

$Z = -0.39$ has a pvalue of 0.348

0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

C) At least 41,000 miles

This is 1 subtracted by the pvalue of Z when X = 41,000. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{41000 - 51200}{8200}$

$Z = -1.24$

$Z = -1.24$ has a pvalue of 0.108

1 - 0.108 = 0.892

0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

D) A lifetime that is within 10,000 miles of the mean

This is the pvalue of Z when X = 51200 + 10000 = 61200 subtracted by the pvalue of Z when X = 51200 - 10000 = 412000. So

X = 61200

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{61200 - 51200}{8200}$

$Z = 1.22$

$Z = 1.22$ has a pvalue of 0.889

X = 41200

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{41200 - 51200}{8200}$

$Z = -1.22$

$Z = -1.22$ has a pvalue of 0.111

0.889 - 0.111 = 0.778

0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

4 0

3 years ago

The cube of the sum of 4 and 9 times x divided by the product of 5 times x and the difference of x and 1

Burka [1]

+. Add x to y x + y y added to 7. 7+ y. The sum of a and b a + b m more than n ... 9 plus m. 9 + m. Subtraction. Subtract, subtract from, difference, between, less, less ... 7 times y. 7y. The product of x and y xy. 5 multiplied by y. 5y one-fifth of p. 1. 5 p ... Power. The square of y y. 2. The cube of k k. 3 t raised to the fourth power t. 4.

5 0

3 years ago