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natita [175]
3 years ago
9

6. The senior class at High School A and High School B planned trips to

Mathematics
1 answer:
Colt1911 [192]3 years ago
4 0
Bus can carry 59 van can carry 18
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CAN SOMEONE PLZ ANSWER THIS ITS DUE AT 11:59 APRIL 11 2021 AND ILL MARK U BRAINLIST
Luden [163]

Answer:

answer is c

Step-by-step explanation:

6 0
2 years ago
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Factorise completely (2x+1)^2 -(x-1)^2
Vladimir79 [104]
Use photomath ! It's an app that gives you the answer with the explanation and it shows you how it got to the answer !
5 0
2 years ago
Nicolas has $650 to deposit into two different savings accounts. • Nicolas will deposit $400 into Account I, which earns 3.5% an
OLga [1]

account 1:

$400 * 0.035 * 2 = $28

$28 interest earned so total = $400 + $28 = $428

account 2:

<span>A = P(1+r/n)^(nt)</span>

= $250(1+0.035)^2 = 267.50

total account 1 and account 2 =

$428 + $267.81 = $695.50

answer is B $695.00


4 0
3 years ago
Find the product of each sum and difference. (2l + 9)(2l – 9)
Alex787 [66]

Answer:

360

Step-by-step explanation:

(i think that is a 1 not an L)

21 + 9 equals 30. And 21-9 equals 12. So you would multiply 30*12 which gives you your answer!

<h2>360</h2>
4 0
2 years ago
A Web ad can be designed from four different colors, three font types, five font sizes, three images, and five text phrases. A s
Varvara68 [4.7K]

Answer:

P(A∪B)=17/20 or 0.85

P(A∪B')=2/5 or 0.4

P(A'∪B')=4/5 or 0.8

Step-by-step explanation:

There are four font colors so each color had equal chance and thus,

P(A)=1/4

There are 5 font sizes and so not the smallest fonts are 4.Thus,

P(B)=4/5

P(A∪B)=P(A)+P(B)-P(A∩B)

The design is generated randomly so event A and event B are independent.

P(A∩B)=P(A)*P(B)

P(A∩B)=1/4(4/5)=1/5

P(A∪B)=P(A)+P(B)-P(A∩B)

P(A∪B)=1/4+4/5-1/5=1/4+3/5

P(A∪B)=17/20 or 0.85

P(A∪B')=P(A)+P(B')-P(A∩B')

P(B')=1-P(B)=1-4/5=1/5

P(A∩B')=P(A)*P(B')=1/4*1/5=1/20

P(A∪B')=P(A)+P(B')-P(A∩B')

P(A∪B')=1/4+1/5-1/20=9/20-1/20=8/20

P(A∪B')=2/5 or 0.4

P(A'∪B')=P(A∩B)'

P(A'∪B')=1-P(A∩B)

P(A'∪B')=1-1/5=4/5

P(A'∪B')=4/5 or 0.8

5 0
3 years ago
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