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3 years ago

Plz helpfind the value of x

Mathematics

1 answer:

gayaneshka [121]3 years ago

6 0

Answer:

$x^{\circ}=70^{\circ}$

Step-by-step explanation:

Lines MR and PQ are parallel. Continue the line MR up to the intersection with the line NQ. Denote the point of their intersection as A.

Consider triangle AMN. In this triangle,

$m\angle ANM=40^{\circ}$ - given;
$m\angle NMA=180^{\circ }-150^{\circ}=30^{\circ}$ - angles RMN and NMA are supplementary angles;

So, (as the sum of all interior angles add up to 180°)

$m\angle NAM=180^{\circ}-40^{\circ}-30^{\circ}=110^{\circ}$

Angles NAM and MAQ are supplementary angles, then

$m\angle MAQ = 180^{\circ}-110^{\circ}=70^{\circ}$

Angles MAQ and AQP are alternate interior angles, so they are congruant and

$x^{\circ}=70^{\circ}$

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In a certain community, 36 percent of the families own a dog and 22 percent of the families that own a dog also own a cat. In ad

zysi [14]

Answer: a) 0.0792 b) 0.264

Step-by-step explanation:

Let Event D = Families own a dog .

Event C = families own a cat .

Given : Probability that families own a dog : P(D)=0.36

Probability that families own a dog also own a cat : P(C|D)=0.22

Probability that families own a cat : P(C)= 0.30

a) Formula to find conditional probability :

$P(B|A)=\dfrac{P(A\cap B)}{P(A)}\\\\\Rightarrow P(A\cap B)=P(B|A)\times P(A)$ (1)

Similarly ,

$P(C\cap D)=P(C|D)\times P(D)\\\\=0.22\times0.36=0.0792$

Hence, the probability that a randomly selected family owns both a dog and a cat : 0.0792

b) Again, using (2)

$P(D|C)=\dfrac{P(C\cap D)}{P(C)}\\\\=\dfrac{0.0792}{0.30}=0.264$

Hence, the conditional probability that a randomly selected family owns a dog given that it owns a cat = 0.264

5 0

3 years ago

Y/a −b= y/b −a, if a≠b solve for y

Kamila [148]

Answer:

y= ab if a≠b

Step-by-step explanation:

y/a −b= y/b −a

multiply each side by ab to clear the fractions

ab(y/a −b) = ab( y/b −a)

distribute

ab * y/a - ab*b = ab * y/b - ab *a

b*y - ab^2 = ay -a^2 b

subtract ay on each side

by -ay -ab^2 = ay-ay -a^2b

by -ay -ab^2 =-a^2b

add ab^2 to each side

by-ay -ab^2 +ab^2 = ab^2 - a^2b

by-ay = ab^2 - a^2b

factor out the y on the left, factor out an ab on the right

y (b-a) = ab(b-a)

divide by (b-a)

y (b-a) /(b-a)= ab(b-a)/(b-a) b-a ≠0 or b≠a

y = ab

5 0

3 years ago

2) Prove that these two lines are parallel:

borishaifa [10]

Answer:

if the two lines have the same slope that means they are parallel

3 0

3 years ago

A.45 B.30 C.90 D.180 E.40 F.55 G.130 H.50

cupoosta [38]

Answer:

If you were to make a 360 degree circle and minus the 50 thats already there you get 310 the right angle on angle 5 is 90 degrees because its a right angle so then the 310-90=220 the obtuse angle on angle 2 is 180 so 220-180=40 lastly you share those forty between angles 3 and 4 and the solution is your answer.

Glab to help!

7 0

3 years ago

The body temperatures of adults are normally distributed with a mean of 98.6degrees° F and a standard deviation of 0.60degrees°

Schach [20]

Answer:

97.72% probability that their mean body temperature is greater than 98.4degrees° F.

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples can be 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$