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3 years ago

Solve for X enter your answer in the box

Mathematics

2 answers:

aliina [53]3 years ago

6 0

Answer:

x = 6

Step-by-step explanation:

san4es73 [151]3 years ago

3 0

Answer:

x = 6

Step-by-step explanation:

GIVEN :-

AD ║ BC
AE = 16 units
AB = 5x + 2 units
DE = 12 units
DC = 24 units

TO FIND :-

Value of x

GENERAL CONCEPT TO BE USED IN THIS QUESTION :-

Basic Proportionality Theorem -

If a line drawn parallel to one side of a triangle and intersects the other two sides at distinct points , it divides the other two sides in equal proportion.

SOLUTION :-

AD is the line drawn parallel to BC and intersects EB & EC at A & D respectively. By using Basic Proportionality Theorem , it can be stated that -

$\frac{AE}{AB} = \frac{DE}{DC}$

$=> \frac{16}{5x + 2} = \frac{12}{24}$

$=> \frac{16}{5x + 2} = \frac{1}{2}$

$=> 5x + 2 = 16 \times 2 = 32$

$=> 5x = 32 - 2 = 30$

$=> x = \frac{30}{5} = 6$

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Let X denote the length of human pregnancies from conception to birth, where X has a normal distribution with mean of 264 days a

Kaylis [27]

Answer:

Step-by-step explanation:

Hello!

X: length of human pregnancies from conception to birth.

X~N(μ;σ²)

μ= 264 day

σ= 16 day

If the variable of interest has a normal distribution, it's the sample mean, that it is also a variable on its own, has a normal distribution with parameters:

X[bar] ~N(μ;σ²/n)

When calculating a probability of a value of "X" happening it corresponds to use the standard normal: Z= (X[bar]-μ)/σ

When calculating the probability of the sample mean taking a given value, the variance is divided by the sample size. The standard normal distribution to use is Z= (X[bar]-μ)/(σ/√n)

a. You need to calculate the probability that the sample mean will be less than 260 for a random sample of 15 women.

P(X[bar]<260)= P(Z<(260-264)/(16/√15))= P(Z<-0.97)= 0.16602

b. P(X[bar]>b)= 0.05

You need to find the value of X[bar] that has above it 5% of the distribution and 95% below.

P(X[bar]≤b)= 0.95

P(Z≤(b-μ)/(σ/√n))= 0.95

The value of Z that accumulates 0.95 of probability is Z= 1.648

Now we reverse the standardization to reach the value of pregnancy length:

1.648= (b-264)/(16/√15)

1.648*(16/√15)= b-264

b= [1.648*(16/√15)]+264

b= 270.81 days

c. Now the sample taken is of 7 women and you need to calculate the probability of the sample mean of the length of pregnancy lies between 1800 and 1900 days.

Symbolically:

P(1800≤X[bar]≤1900) = P(X[bar]≤1900) - P(X[bar]≤1800)

P(Z≤(1900-264)/(16/√7)) - P(Z≤(1800-264)/(16/√7))

P(Z≤270.53) - P(Z≤253.99)= 1 - 1 = 0

d. P(X[bar]>270)= 0.1151

P(Z>(270-264)/(16/√n))= 0.1151

P(Z≤(270-264)/(16/√n))= 1 - 0.1151

P(Z≤6/(16/√n))= 0.8849

With the information of the cumulated probability you can reach the value of Z and clear the sample size needed:

P(Z≤1.200)= 0.8849

$Z= \frac{X[bar]-Mu}{Sigma/\sqrt{n} }$