All categories

3 years ago

A. 115 B. 167 C. 126 D. 96

Mathematics

2 answers:

geniusboy [140]3 years ago

5 0

Answer:

126

Step-by-step explanation:

Let x be the missing length

The triangles are similar:

● UE/140 = 45/x

From the graph we deduce that:

● UE = 140 - 90 = 50

Replace UE by its value

● 50/ 140 = 45/x

Switch x and 50

● x / 140 = 45/50

45/50 is 9/10 wich is 0.9

● x/140 = 0.9

Multiply 0.9 by 140

● x = 140 × 0.9

● x = 126

Bumek [7]3 years ago

3 0

Answer:

I think its c 126

Step-by-step explanation:

You might be interested in

What is the y-intercept from the point (0,7)

blsea [12.9K]

Answer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Let U be the universal set, where U=(1,2,3,4,5,6,7,8,9)let sets A and B. be subsets of U, whereA=(2,3,8,9) and B=(1,2,3,5)

Black_prince [1.1K]

(A U B) = {1, 2, 3, 5, 8, 9}

(A U B)^c = {4, 6, 7}

A^c = {1, 4, 5, 6, 7 }

B^c = {4, 6, 7, 8, 9}

A^c ∩ B^c = {4, 6, 7}

8 + 3(4 + 5)

8 + 3(9)

8 + 27 = 35

6 0

1 year ago

The table shows the results for spinning the spinner 75 times. What is the relative frequency for the event "spin a 3"?

Vadim26 [7]

Answer: 0.267

Step-by-step explanation:

The relative frequency is the number of times that we obtained a given outcome divided by the total numer of trials.

In this case we have 75 trials.

in 20 trials the outcome was a 3.

Then the relative frequency of the event "spin a 3" is:

p = 20/75 = 0.266... = 0.267

3 0

3 years ago

Consider the following data for 120 mathematics students at a college concerning the languages French, German, and Russian:

Vlada [557]

Answer:

check online for more information

7 0

3 years ago

The age of the children in kindergarten on the first day of school is uniformly distributed between 4.8 and 5.8 years old. A fir

Kazeer [188]

Answer:

(1) (c) 5.30 years.

(2) (b) 0.289.

(3) (b) 0.80.

(4) (d) 0.50.

(5) (a) 5.25 years.

Step-by-step explanation:

Let X = age of the children in kindergarten on the first day of school.

The random variable X follows a continuous Uniform distribution with parameters a = 4.8 years and b = 5.8 years.

The probability density function function of X is:

$f_{X}(x)=\left \{ {{\frac{1}{b-a}} ;\ a$

(1)

The expected value of a Uniform random variable is:

$E(X)=\frac{1}{2}(a+b)$

Compute the mean of X as follows:

$E(X)=\frac{1}{2}(a+b)=\frac{1}{2}\times (4.8+5.8)=5.3$

Thus, the mean of the distribution is (c) 5.30 years.

(2)

The standard deviation of a Uniform random variable is:

$SD(X)=\sqrt{\frac{1}{12}(b-a)^{2}}$

Compute the standard deviation of X as follows:

$SD(X)=\sqrt{\frac{1}{12}(b-a)^{2}}=\sqrt{\frac{1}{12}\times (5.8-4.8)^{2}}=0.289$

Thus, the standard deviation of the distribution is (b) 0.289.

(3)

Compute the probability that a randomly selected child is older than 5 years old as follows:

$P(X>5)=\int\limits^{5.8}_{5} {\frac{1}{5.8-4.8}}\, dx\\$

$=\int\limits^{5.8}_{5} {1}\, dx\\=[x]^{5.8}_{5}\\=(5.8-5)\\=0.8$

Thus, the probability that a randomly selected child is older than 5 years old is (b) 0.80.

(4)

Compute the probability that a randomly selected child is between 5.2 years and 5.7 years old as follows:

$P(5.2$

$=\int\limits^{5.7}_{5.2} {1}\, dx\\=[x]^{5.7}_{5.2}\\=(5.7-5.2)\\=0.5$

Thus, the probability that a randomly selected child is between 5.2 years and 5.7 years old is (d) 0.50.

(5)

It is provided that a randomly selected child is at the 45th percentile.

This implies that:

P (X < x) = 0.45

Compute the value of x as follows:

$P (X < x) = 0.45$

$\int\limits^{x}_{4.8} {\frac{1}{5.8-4.8}}\, dx=0.45$

$\int\limits^{x}_{4.8} {1}\, dx=0.45$

$[x]^{x}_{4.8}=0.45$

$x-4.8=0.45\\$

$x=0.45+4.8\\x=5.25$

Thus, the age of the child at the 45th percentile is (a) 5.25 years.

6 0

3 years ago