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Orlov [11]
3 years ago
8

How much time Eight can you go into 400 and how much time 5 in going to 64 and how much time 4 can going into to 78

Mathematics
1 answer:
Ann [662]3 years ago
8 0

Answer:

Step-by-step explanation:

400/8=50

64/5=12.8

78/4=39/2=19.5

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Step-by-step explanation:

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Ronald bought a number of cds for 13$ each. What was the total amount he spent on CDs
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How many cds did he buy?
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Choose the equation that represents a line that passes through points (−1, 2) and (3, 1).
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You did not provide us with equations to select.

Find the slope m.

m = (1 - 2)/(3 - (-1))

m = -1/(3 + 1)

m = -1/4

Use the slope and one of the points and plug into the point-slope formula.

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5 0
3 years ago
Read 2 more answers
A new security system needs to be evaluated in the airport. The probability of a person being a security hazard is 4%. At the ch
ivolga24 [154]

Answer:

(a) 0.9412

(b) 0.9996 ≈ 1

Step-by-step explanation:

Denote the events a follows:

P = a person passes the security system

H = a person is a security hazard

Given:

P (H) = 0.04,\ P(P^{c}|H^{c})=0.02\ and\ P(P|H)=0.01

Then,

P(H^{c})=1-P(H)=1-0.04=0.96\\P(P|H^{c})=1-P(P|H)=1-0.02=0.98\\

(a)

Compute the probability that a person passes the security system using the total probability rule as follows:

The total probability rule states that: P(A)=P(A|B)P(B)+P(A|B^{c})P(B^{c})

The value of P (P) is:

P(P)=P(P|H)P(H)+P(P|H^{c})P(H^{c})\\=(0.01\times0.04)+(0.98\times0.96)\\=0.9412

Thus, the probability that a person passes the security system is 0.9412.

(b)

Compute the probability that a person who passes through the system is without any security problems as follows:

P(H^{c}|P)=\frac{P(P|H^{c})P(H^{c})}{P(P)} \\=\frac{0.98\times0.96}{0.9412} \\=0.9996\\\approx1

Thus, the probability that a person who passes through the system is without any security problems is approximately 1.

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3 years ago
The house numbers on Carr Memorial Avenue follow a pattern. The first four houses on the left side of the street are numbered 8,
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There would be 4 more houses on the left side of the street
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