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3 years ago

28) (-5ưyA +9u) + (-5ư v4 - 8u + 8u²v2)+(-8u*v + 8uº4)=

Mathematics

1 answer:

Mice21 [21]3 years ago

4 0

Answer: good luck

Step-by-step explanation:

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What is the solution to the equation below? 4+√x=10 A. 196 B. 36 C. 6 D. 12

Helga [31]

The Answer to Your Question is B. 36

Be sure to give me Brainliest Answer... Helps alot :) THX

3 0

3 years ago

Read 2 more answers

wrote an equation for the line. passing through (-5,7) and parallel to the line whise equation is 5x-3y-4=0

xeze [42]

Step-by-step explanation:

5x-3y-4=0---------(1)

Slope = - coefficient of X/ coefficient of y

= -5/(-3)

= 5/3

(x1,y1)= (-5,7)

Equation of the line,

Y-Y1= slope(X-X1)

Y-7= (5/3)×(X-(-5))

Y-7= (5x+25)/3

3y-7=5x+25

5x-3y+32=0 is the required eqn

3 0

3 years ago

What is the mean absolute deviation for the data? 20,25,30,30,45

Ivan

Add them all together the divide what you get by 5

6 0

3 years ago

Rosa works due north of home. Her husband Juan works due east. They leave for work at the same time. By the time Rosa is 7 miles

polet [3.4K]

This is a right triangle problem. c^2 = a^2 + b^2
Let b = 7 mi (Rosa's distance from home)
Let a = the east leg (Juan's distance from home
(a+1) = c (hypotenuse, distance between them)

(a+1)^2 = a^2 + 7^2

FOIL (a+1)(a+1)
a^2 + 2a + 1 = a^2 + 49

Combine like terms
a^2 - a^2 + 2a = 49 - 1
2a = 48
a =

a = 24 mi, Juan's distance from home

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

8 0

2 years ago

he amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and s

sp2606 [1]

Complete question:

He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is

a) less than 8 minutes

b) between 8 and 9 minutes

c) less than 7.5 minutes

Answer:

a) 0.0708

b) 0.9291

c) 0.0000

Step-by-step explanation:

Given:

n = 47

u = 8.3 mins

s.d = 1.4 mins

a) Less than 8 minutes:

$P(X$

P(X' < 8) = P(Z< - 1.47)

Using the normal distribution table:

NORMSDIST(-1.47)

= 0.0708

b) between 8 and 9 minutes:

P(8< X' <9) = $[\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}]$

= P(-1.47 <Z< 6.366)

= P( Z< 6.366) - P(Z< -1.47)

Using normal distribution table,

$NORMSDIST(6.366)-NORMSDIST(-1.47)$

0.9999 - 0.0708

= 0.9291

c) Less than 7.5 minutes:

P(X'<7.5) = $P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}]$

P(X' < 7.5) = P(Z< -3.92)

NORMSDIST (-3.92)

= 0.0000

3 0

2 years ago