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4 years ago

Identify the length of BP. PLEASE HELP ASAP!!

Mathematics

2 answers:

Marianna [84]4 years ago

4 0

Answer:

15.75

Step-by-step explanation:

iogann1982 [59]4 years ago

3 0

Answer:

15.75

Step-by-step explanation:

If you look at this as a trapezoid, the ratio of the sides AH to PT is 3:2, so just use this ratio for the other parts. 10.5*3/2=15.75

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5.2x-y=4.1 1.5x+y=6.7<br> how to solve this problem

MrRa [10]

Answer: x=1.6

y=4.2

Step-by-step explanation:

5.2x - y = 4.1

1.5x + y = 6.7

we will add both equations and you see that the y is getting cancelled

6.7x=10.8

x=10.8/6.7= 1.6

now in any equation we will replace x with 1.6

5.2(1.6)-y=4.1

8.3-y=4.1

-y= 4.1-8.3

-y=-4.2 multiply by -1

y=4.2

8 0

3 years ago

Read 2 more answers

Which expression is equivalent to 42+90?

MrRissso [65]

Step-by-step explanation:

42+90 can be written in many ways.

Fore example, it can be written with double negatives, or even commutative property.

42-(-90) or 90+42

(21x2)+(45x2) or (6x7)+(3x30)

Sorry if it's wrong...

4 0

4 years ago

How do I solve this step by step?

Mashcka [7]

First step : 2(2))+5\2 -2(2))+3\2 =9\2 -7\2 =2\2 =1

5 0

3 years ago

4c+5 = 8c-11<br>what is the variable

ozzi

C the variable equals 4

8 0

4 years ago

Read 2 more answers

A study was conducted to measure the effectiveness of hypnotism in reducing pain. The measurements are centimeters on a pain sca

inna [77]

Answer:

Step-by-step explanation:

Hello!

This is an example of a pared sample test, the experiment is based on two dependent variables:

X₁: centimeters on a pain scale before hypnosis

X₂: centimeters on a pain scale after hypnosis

Out of these two variables a new variable is determined Xd= X₁-X₂

If the variables have an approximate normal distribution then the variable resulting from their difference will also have an approximate normal distribution.

The claim is that "hypnosis reduced the pain" if so you'd expect the population mean of the difference to be less than zero, symbolically: μd<0

The statistic for this test is a paired sample t test:

$t= \frac{\frac{}{X_d} - Mu_d}{Sd} ~t_{n-1}$

To calculate the sample mean and variance you have to calculate the difference between the pairs first.

$\frac{}{Xd}$ = ∑Dif/n

$S_d^2= \frac{1}{n-1} [sumDif^2- \frac{(sumDif)^2}{n} ]$

∑Dif= 6.4

∑Dif²= 12.64

$\frac{}{Xd}$ = 6.4/5= 1.28

$S_d^2= \frac{1}{4} [12.64- \frac{(6.4)^2}{5} ]= 1.112$

Sd= 1.05

$t_{H_0}= \frac{\frac{}{Xd}-Mu_d }{Sd} = \frac{1.28-0}{1.05} = 1.219= 1.22$

I hope this helps!

7 0

4 years ago