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Lana71 [14]
3 years ago
12

How many seven digit telephone numbers are possible the first digit cannot be three and the number must be a multiple of 10

Mathematics
1 answer:
Korolek [52]3 years ago
7 0

Answer:

900,000.

Step-by-step explanation:

If the number is a multiple of 10 it must end in zero.

The first digit can be any one of 0-9 except 3 - that is one of 9 digits.

All other positions (2nd to 6th) can be any one of 0 to 9 (10 digits).

So the number of possible telephone numbers is:

9*10*10*10*10*10

= 900,000.

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Shkiper50 [21]

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3 years ago
Help me pls. Given the slope is 1/3, the y-intercept is 4, and a point of the graph is (3,5). Write the equations of the linear
enyata [817]
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3 years ago
I. If 40% of all commuters ride to work in carpools, if eight workers are selected, find the probability that exactly five will
snow_tiger [21]

Complete question is;

I. If 40% of all commuters ride to work in carpools, if eight workers are selected, find the probability that exactly five will ride in carpools.

II. Traffic lights with a turning arrow are red 70% of the time. If you approach 6 lights in a row, find the probability of stopping: a. Exactly four times. b. At least three times.

Answer:

I)P(X = 5) = 0.1239

IIa)P(X = 4) = 0.3241

IIb)P(X ≥ 3) = 0.9294

Step-by-step explanation:

I) This is a binomial probability problem.

Formula is;

P(X = x) = nCx(p^(x) × (1 - p)^(n - x)

40% of all commuters ride to work in carpools. Thus, p = 0.4

eight workers are selected and we are to find the probability that exactly five will ride in carpools.

Thus, n = 8 and x = 5

P(X = 5) = 8C5 × 0.4^(5) × (1 - 0.4)^(8 - 5)

P(X = 5) = 56 × 0.01024 × 0.6³

P(X = 5) = 0.1239

II) Traffic lights with a turning arrow are red 70% of the time and red means stop.

Thus; p = 0.7

Since 6 lights in a row;

n = 6

a) Probability of stopping exactly 4 times;

P(X = 4) = 6C4 × 0.7⁴ × (1 - 0.7)^(6 - 4)

P(X = 4) = 15 × 0.2401 × 0.09

P(X = 4) = 0.3241

b) probability of at least 3 times is;

P(X ≥ 3) = P(3) + P(4) + P(5) + P(6)

P(3) = 6C3 × 0.7³ × (1 - 0.7)^(6 - 3)

P(3) = 0.1852

P(4) = 0.3241

P(5) = 6C5 × 0.7^(5) × (1 - 0.7)^(6 - 5)

P(5) = 0.3025

P(6) = 6C6 × 0.7^(6) × (1 - 0.7)^(6 - 6)

P(6) = 0.1176

P(X ≥ 3) = 0.1852 + 0.3241 + 0.3025 + 0.1176

P(X ≥ 3) = 0.9294

5 0
3 years ago
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