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3 years ago

Examine parallelogram ABCD below.

Mathematics

1 answer:

gtnhenbr [62]3 years ago

6 0

Given that,

∠C = (6x-21)° and ∠A = (4x+11)°

To find,

Choose the correct option.

Solution,

For a parallelogram, the opposite angles are equal. ATQ,

(6x-21)° = (4x+11)°

Taking like terms together,

6x-4x = 21 + 11

2x = 32

x = 16

∠C = (6x-21)° = 6(16) -21 = 75°

So, ∠A = 75°

So, ∠A = ∠C = 75°. Hence, the correct option is (B).

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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

3 years ago

(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2

Montano1993 [528]

Answer:

$\dfrac{m^3}{16p^2q^2}$

Step-by-step explanation:

Given:

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}$

$m^{-1}=\dfrac{1}{m}$

$q^0=1$

$2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}$

$(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}$

$m^{-1}=\dfrac{1}{m}$

$2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}$

$\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}$

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}$

8 0

3 years ago

PLEASE HELP ME ON THIS!!!!!!!ASAP!!!!!!!!!!!!!!!!!!

NemiM [27]

Answer:

B is the right answer

Step-by-step explanation:

because of negative 7 on the right it shows that it is down 7

8 0

3 years ago

Please help me <br>ineed it now

yanalaym [24]

Answer:

look at explanation

Step-by-step explanation:

To solve all of these questions you just need to scale. While scaling you can divide or multiply.

1. $\frac{180 km}{4 hrs}$ (divide by 4 to get unit rate)

unit rate = $\frac{45km}{1hr}$

2. If it takes one hour to travel 40 miles, then we don't have to scale or anything. We know $\frac{40 miles}{1 hr}$ , the answer is...

"It'll only take one hour to travel 40 miles. This is correct because the unit rate is $\frac{40 miles}{1 hr}$ . "

3. Knowing that the constant speed of the plane is $\frac{800 km}{1 hour\\}$ , we can scale then add half of the unit rate to get our answer.

$\frac{800 km}{1 hour\\}$ x 3 = $\frac{2400 km}{3 hrs}$ 800 x 3 = 2400

then add half of unit rate which is $\frac{400 km}{30 mins}$ $\sqrt[2]{800}$ = 400

so, our answer is $\frac{2800 km}{3.5 hours}$

4. $\frac{3km}{30 mins}$ <--- ( divided by 2) $\frac{6 km}{1 hr}$ ( multiplied x 2.5) ---> $\frac{15 km}{2.5 hrs}$

To get an easier way to find the answer, if possible you can scale back farther than the 1 hour mark.

Elmer would take 2 and a half hours to ride 15 km.

5. The boys speed is $\frac{4 km}{1 hr}$ . All we need to do is divide by 2. $\sqrt[2]{8}$ = 4

3 0

3 years ago

A rectangular prism has a length of 9 centimeters, a width of 4 centimeters, and a height of 2 centimeters.

RUDIKE [14]

Answer:

It will be 72cm

Step-by-step explanation:

Volume of a rectangular prism = (length x width x height) cubic units.

9x4x2=72

5 0

3 years ago