(15x-7)-(23x-2) use foil

JKLM is a rhombus. KM is 10 and JL is 24. Find the perimeter of the rhombus.

Ahat [919]

Given that JKLM is a rhombus and the length of diagonal KM=10 na d JL=24, the perimeter will be found as follows;
the length of one side of the rhombus will be given by Pythagorean theorem, the reason being at the point the diagonals intersect, they form a perpendicular angles;
thus
c^2=a^2+b^2
hence;
c^2=5^2+12^2
c^2=144+25
c^2=169
thus;
c=sqrt169
c=13 units;
thus the perimeter of the rhombus will be:
P=L+L+L+L
P=13+13+13+13
P=52 units

5 0

3 years ago

Assume that females have pulse rates that are normally distributed with a mean of μ=73.0 beats per minute and a standard deviati

Gennadij [26K]

Answer:

a. the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly selected, the probability that they have pulse rates with a mean less than 76 beats per minute is 0.8849

c. D. Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

Given that:

Mean μ =73.0

Standard deviation σ =12.5

a. If 1 adult female is randomly selected, find the probability that her pulse rate is less than 76 beats per minute.

Let X represent the random variable that is normally distributed with a mean of 73.0 beats per minute and a standard deviation of 12.5 beats per minute.

Then : X $\sim$ N ( μ = 73.0 , σ = 12.5)

The probability that her pulse rate is less than 76 beats per minute can be computed as:

$P(X < 76) = P(\dfrac{X-\mu}{\sigma}< \dfrac{X-\mu}{\sigma})$

$P(X < 76) = P(\dfrac{76-\mu}{\sigma}< \dfrac{76-73}{12.5})$

$P(X < 76) = P(Z< \dfrac{3}{12.5})$

$P(X < 76) = P(Z< 0.24)$

From the standard normal distribution tables,

$P(X < 76) = 0.5948$

Therefore , the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly selected, find the probability that they have pulse rates with a mean less than 76 beats per minute.

now; we have a sample size n = 25

The probability can now be calculated as follows:

$P(\overline X < 76) = P(\dfrac{\overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{ \overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}})$

$P( \overline X < 76) = P(\dfrac{76-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{76-73}{\dfrac{12.5}{\sqrt{25}}})$

$P( \overline X < 76) = P(Z< \dfrac{3}{\dfrac{12.5}{5}})$

$P( \overline X < 76) = P(Z< 1.2)$

From the standard normal distribution tables,

$P(\overline X < 76) = 0.8849$

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

In order to determine the probability in part (b); the normal distribution is perfect to be used here even when the sample size does not exceed 30.

Therefore option D is correct.

Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

5 0

2 years ago

Solve For Q q/10=4 Please Answer This For Me. Thanks.

Maurinko [17]

$\frac{q}{10} = 4\\q = 4 * 10\\q = 40$

Answer: 40.

3 0

3 years ago

Read 2 more answers

Find the measure of the angle labeled x (ADB) in the circle below. A B

My name is Ann [436]

The answer is the last option!
17’

8 0

2 years ago

Solve: $|2x - 1| = |3x + 5|$. write your answers as a list of numbers, separated by commas,

emmainna [20.7K]

First, disregard the sign for absolute value and solve for x.

2x - 1 = 3x + 5
2x - 3x = 5 + 1
-x = 6
x = -6

Now, you have to interpret the absolute value. The absolute value of x is always the positive version of whatever the number is. Since x = -6, its absolute value is 6. Therefore, the possible value for x is only one, which is 6.

8 0

3 years ago

(15x-7)-(23x-2)use foil​

(15x-7)-(23x-2)use foil