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3 years ago

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Mathematics

1 answer:

weeeeeb [17]3 years ago

5 0

Answer:

width = 2v - 19

Step-by-step explanation:

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Make x the subject of the formula a(x+b)=c pls help meeeee

velikii [3]

Answer: c(a+ab)=x

Step-by-step explanation:

a(x+b)=c distribute the (a)

ax+ab=c now divide by ax by x

a+ab=c/x now multiply by c

c(a+ab)=x

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4 years ago

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2x+y=-2 5x+3y=-8 Solve the following systems

AlekseyPX

This is the first step

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3 years ago

Can someone help me do this?

tester [92]

A. -13ab

B. 5a^2b

C. 3

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3 years ago

PLS HELP WILL MARK BRAINLIEST!! TY

Simora [160]

Step-by-step explanation:

$\angle{1} = 77°$ because they are vertical angles.

The 77°-angle is supplementary with $\angle{2}$ therefore

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$\angle{3}$ and $\angle{2}$ are vertical angles so

$\angle{3} = \angle{2} = 103°$

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3 years ago

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Textbook prices have a seasonal structure on Ebay. At the end of a term, the supply of used books outstrips demand, and the pric

TiliK225 [7]

Answer:

(1)$68.02

(2)

a.) $P(x\geq 2)=0.76$

b.) $P(x\geq 3)=0.5$

c.) $P(x> 3)=0.38$

d.) $P(x< 2)=0.24$

e.) $P(x\leq 2)=0.5$

f) $\mu=2.72$

g) $\sigma=1.33$

Step-by-step explanation:

Question 1

The probability distribution table for the average price of the textbook sales in each of these three time periods is given below:

$\left|\begin{array}{c|c|c|c}&$Start-of-term&$End-of-term&$Other\\$Sales proportion&0.45&0.31&0.24\\$Average price&\$82.52&\$49.12&\$65.23\end{array}\right|$

We are required to calculate the average price of the textbook over all seasons.

Expected Value

$= (0.45 X 82.52)+(0.31 X 49.12) +(0.24 X 65.23)\\=68.02$

The average price of the textbook over all seasons is $68.02

Question 2

Distribution Table for Number of Courses being taken by BHCC Students

$\left|\begin{array}{c|ccccc}x$(No of classes)& 1&2&3 &4&5\\P(x)&0.24&0.26&0.12&0.3 &0.08\end{array}\right|$

a.) Probability that a student is taking 2 or more classes

$P(x\geq 2)=0.26+0.12+0.3+0.08=0.76$

b.) Probability that a student is taking at least 3 classes

$P(x\geq 3)=0.12+0.3+0.08=0.5$

c.) Probability that a student is taking more than 3 classes

$P(x> 3)=0.3+0.08=0.38$

d.) Probability that a student is taking less than 2 classes

$P(x< 2)=0.24$

e.) Probability that a student is taking no more than 2 classes

$P(x\leq 2)=0.24+0.26=0.5$

f)Average (mean) amount of classes

$=(1*0.24)+(2*0.26)+(3*0.12)+(4*0.3)+(5*0.08)\\\mu=2.72$

g)Standard deviation for the amount of classes

$\left|\begin{array}{c|ccccc|c}x$(No of classes)& 1&2&3 &4&5&Sum\\x-\mu&-1.72&-0.72&0.28&1.28&2.28\\(x-\mu)^2&2.9584&0.5184&0.0784&1.6384&5.1984\\P(x)&0.24&0.26&0.12&0.3 &0.08\\--&--&---&---&--&--&--\\(x-\mu)^2P(x)&0.71&0.13&0.01&0.49&0.42&1.76\end{array}\right|$

Standard Deviation

$=\sqrt{\sum(x-\mu)^2P(x)} \\=\sqrt{1.76} \\\sigma =1.33$

7 0

3 years ago