3. Suppose that this statement is true: If I wear boots

Can you help me find the value of x and y. I am very confused.

Monica [59]

Hello,
Let's assume top left corner: A
top right corner : B
Bottom right corner: C
Bottom left corner :D

M= middle of [CD]

ABM is a triangle rectangular isocel:

(3√2)²+(3√2)²=y²
==>y²=2*9*2
==>y²=36
==>y=6

The triangle BCM is rectangular with MB=3√2, MC=y/2=3
x²+3²=(3√2)²
==>x²=9*2-9
==>x²=9
==>x=3

8 0

3 years ago

Please please help me!!!!!

Tamiku [17]

Your just multiplying 7 and 10, so your answer is gonna be B.

6 0

3 years ago

The two legs of a right triangle are 20 and x - 8, and the hypotenuse is x. Find the length of the hypotenuse.

oksian1 [2.3K]

20^2 + (x-8)^2 = x^2

x^2 = 400 + x^2 -16x + 64

16x = 464

x = 29 = hypotenuse

Double - Check
29^2 = 20^2 + 21^2
841 = 400 + 441

8 0

4 years ago

What is the answer to 2 3/8 ÷ 1 1/4

OLEGan [10]

$2\frac{3}{8}$ ÷ $1\frac{1}{4}$ equals $1\frac{9}{10}$ .

First, convert $2 \frac{3}{8}$ to improper fraction. Use this rule: $a \frac{b}{c}$ = $\frac{ac+b}{c}$ . Your problem should look like: $\frac{2x8+3}{8}$ ÷ $1\frac{1}{4}$ .
Second, simplify 2 x 8 to get 16. Your problem should look like: $\frac{16+3}{8}$ ÷ $1 \frac{1}{4}$ .
Third, simplify 16 + 3 to get 19. Your problem should look like: $\frac{19}{8}$ ÷ $1 \frac{1}{4}$ .
Fourth, convert $1\frac{1}{4}$ to improper fraction. Use the same rule as earlier. Your problem should look like: $\frac{19}{8}$ ÷ $\frac{4+1}{4}$ .
Fifth, simplify 4 + 1 to get 5. Your problem should look like: $\frac{19}{8}$ ÷ $\frac{5}{4}$ .
Sixth, apply this rule: a ÷ $\frac{b}{c}$ = a × $\frac{c}{b}$ . Your problem should look like: $\frac{19}{8}$ × $\frac{4}{5}$ .
Seventh, apply this rule: $\frac{a}{b}$ × $\frac{c}{d}$ = $\frac{ac}{bd}$ . Your problem should look like: $\frac{19x4}{8x5}$ .
Eighth, simplify 19 × 4 to get 76. Your problem should look like: $\frac{76}{8x5}$ .
Ninth, simplify 8 × 5 to get 40. Your problem should look like: $\frac{76}{40}$ .
Tenth, simplify. Your problem should look like: $\frac{19}{10}$ .
Eleventh, convert to mixed fraction. Your problem should look like: $1 \frac{9}{10}$ which is the answer.

Answer as mixed number form: $1\frac{9}{10}$ .
Answer as exact form: $\frac{19}{10}$ .
Answer as decimal form: 1.9.

7 0

3 years ago

Read 2 more answers

The following data represent the pH of rain for a random sample of 12 rain dates. A normal probability plot suggests the data co

Travka [436]

Answer:

a) Point estimate of the population mean = 4.883

b) B.There is 95% confidence that the population mean pH of rain water is between 4.646 and 5.120.

c) C.There is 99% confidence that the population mean pH of rain water is between 4.549 and 5.217.

d) As the level of confidence increases, the width of the interval increases.

This makes sense since the margin of error increases as well.

Step-by-step explanation:

We have a sample for the pH of rain.

The mean of the sample is:

$M=\dfrac{1}{12}\sum_{i=1}^{12}(5.3+5.72+4.38+4.8+5.02+...+4.56+4.68)\\\\\\ M=\dfrac{58.59}{12}=4.883$

The sample standard deviation is:

$s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{12}(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{11}\cdot [(5.3-4.883)^2+(5.72-4.883)^2+...+(4.68-4.883)^2]}\\\\\\$

$s=\sqrt{\dfrac{1}{11}\cdot [(0.17)+...+(0.1)+(0.04)]}\\\\\\s=\sqrt{\dfrac{1.526625}{11}}=\sqrt{0.1387841}\\\\\\s=0.373$

a) The point estimation for the population mean is the sample mean and has a value of 4.883.

b) We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=4.883.

The sample size is N=12.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{0.373}{\sqrt{12}}=\dfrac{0.373}{3.464}=0.108$

The t-value for a 95% confidence interval is t=2.201.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=2.201 \cdot 0.108=0.237$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 4.883-0.237=4.646\\\\UL=M+t \cdot s_M = 4.883+0.237=5.12$

The 95% confidence interval for the mean is (4.646, 5.120).

B.There is 95% confidence that the population mean pH of rain water is between 4.646 and 5.120.

c) We have to calculate a 99% confidence interval for the mean.

The t-value for a 99% confidence interval is t=3.106.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=3.106 \cdot 0.108=0.334$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 4.883-0.334=4.549\\\\UL=M+t \cdot s_M = 4.883+0.334=5.217$

The 99% confidence interval for the mean is (4.549, 5.217).

C.There is 99% confidence that the population mean pH of rain water is between 4.549 and 5.217.

d) When the confidence level is increased, the width also increases as it has to include more possible values for the true mean of the population.

As the level of confidence increases, the width of the interval increases.

This makes sense since the margin of error increases as well.

5 0

4 years ago