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Digiron [165]
3 years ago
11

I need your help plz

Mathematics
1 answer:
vlada-n [284]3 years ago
7 0

Answer:

d = -10/7 = -1 3/7

that is the answer for number 1.

For number 2. the answer is U = 13/4 = 3 1/4 = 3.25

step by step explanation

multiply both sides of the equation by 4 the least common multiple of 2,4

2 (5 x 2 + 1) -4 u = 9 then multiply 5 and 2 to get 10

2 (10 + 1) -4u =9

then add 10 and one to get 11

2 x 11 -4u =9

then multiply two and 11 to get 22

22 -4u = 9

then subtract 22 from both sides

subtract 22 - 9 to get -13

-4u = -13 then undo multiplication by dividing both sides divide both sides by 4-13 / -4 can be divided by positive 13/4

your answer is 13/4

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In​ randomized, double-blind clinical trials of a new​ vaccine, were randomly divided into two groups. Subjects in group 1 recei
Ksivusya [100]

Answer: YES.

the evidence supports this claim.

Step-by-step explanation:

We have that from the complete information,

H0: p1 = p2

Ha : p1 > p2

N1 = 654, P1 = 103/654

N2 = 615, P2 = 52/615

First we have to check the states of typicality that is if n1p1 and n1*(1-p1) and n2*p2 and n2*(1-p2) all are more prominent than equivalent to 5 or not

N1*p1 = 103

N1*(1-p1) = 551

N2*p2 = 52

N2*(1-p2) = 563

All the conditions are met so we can utilize standard typical z table to direct the test

Test measurements z = (P1-P2)/standard mistake

Standard blunder = √{p*(1-p)}*√{(1/n1)+(1/n2)}

P = pooled extent = [(p1*n1)+(p2*n2)]/[n1+n2]

After replacement

Test measurements z = 3.97

From z table, P(Z>3.97) = 0.000036

Thus, P-Value = 0.000036

As the got P-Value is not exactly the given noteworthiness 0.01

We reject the invalid speculation

Thus, we have enough proof to help the claim.that a higher extent of subjects in bunch 1 experienced drowsiness.

cheers I hope this helped !!

5 0
3 years ago
2-61. One of Teddy's jobs at home is to pump gas for his family's sedan and truck. When he fills up the sedan with 12 gallons of
Alborosie

Answer: the answer is 4.19 I found this by dividing 50.28 by 12.   I don't know if this is called the unit

7 0
1 year ago
What is the amplitude, period, and phase shift of y= 5cos( x/2 + 2pi/3)
nordsb [41]
Y = A cos(Bx + C)
A - amplitude
\frac{2\pi}{B} - period
- \frac{C}{B} - phase shift

y=5\cos{( \frac{x}{2} + \frac{2\pi}{3} )}
\\
\\y=5\cos{( \frac{1}{2}x + \frac{2\pi}{3})}
\\
\\A=5,B= \frac{1}{2},C= \frac{2\pi}{3}

Amplitude: 5
Period: \frac{2\pi}{B}=\frac{2\pi}{\frac{1}{2}}=4\pi
Phase shift: - \frac{C}{B} =- \frac{\frac{2\pi}{3}}{\frac{1}{2}}= -\frac{4\pi}{3}
3 0
3 years ago
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2 lines perpendicular to a third line, are parallel to each other, so the tangents t and l are parallel to each other.
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I would say to add the 2 prices together and that would be your answer! 
So your answer is 261.81!
Hope this helps!
5 0
3 years ago
Read 2 more answers
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