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3 years ago

9

Jonah finished a race in 11.62 seconds. Marco finished the same race in 9.8 seconds. How many seconds faster did Marco finish th

e race than Jonah?
A. 1.82 seconds

B. 2.22 seconds

C. 10.64 seconds

D. 11.36 seconds

2 answers:

Nitella [24]3 years ago

6 0

The answer is A. = 1.82 seconds

Lina20 [59]3 years ago

4 0

BBBBBBBBBBBBBBBBBBBbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb

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You always need some time to get up after the alarm has rung. You get up from 10 to 20 minutes later, with any time in that inte

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Answer:

a) P(x<5)=0.

b) E(X)=15.

c) P(8<x<13)=0.3.

d) P=0.216.

e) P=1.

Step-by-step explanation:

We have the function:

$f(x)=\left \{ {{\frac{1}{10},\, \, \, 10\leq x\leq 20 } \atop {0, \, \, \, \, \, \, otherwise }} \right.$

a) We calculate the probability that you need less than 5 minutes to get up:

$P(x$

Therefore, the probability is P(x<5)=0.

b) It takes us between 10 and 20 minutes to get up. The expected value is to get up in 15 minutes.

E(X)=15.

c) We calculate the probability that you will need between 8 and 13 minutes:

$P(8\leq x\leq 13)=P(10\leqx\leq 13)\\\\P(8\leq x\leq 13)=\int_{10}^{13} f(x)\, dx\\\\P(8\leq x\leq 13)=\int_{10}^{13} \frac{1}{10} \, dx\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot [x]_{10}^{13}\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot (13-10)\\\\P(8\leq x\leq 13)=\frac{3}{10}\\\\P(8\leq x\leq 13)=0.3$

Therefore, the probability is P(8<x<13)=0.3.

d) We calculate the probability that you will be late to each of the 9:30am classes next week:

$P(x>14)=\int_{14}^{20} f(x)\, dx\\\\P(x>14)=\int_{14}^{20} \frac{1}{10} \, dx\\\\P(x>14)=\frac{1}{10} [x]_{14}^{20}\\\\P(x>14)=\frac{6}{10}\\\\P(x>14)=0.6$

You have 9:30am classes three times a week. So, we get:

$P=0.6^3=0.216$

Therefore, the probability is P=0.216.

e) We calculate the probability that you are late to at least one 9am class next week:

$P(x>9.5)=\int_{10}^{20} f(x)\, dx\\\\P(x>9.5)=\int_{10}^{20} \frac{1}{10} \, dx\\\\P(x>9.5)=\frac{1}{10} [x]_{10}^{20}\\\\P(x>9.5)=1$

Therefore, the probability is P=1.

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3 years ago

Find the nth term of the arithmetic sequences<br> a1=5,d=6,n=11

ki77a [65]

here's the solution,

n = 11
a = 5 ( a = first term )
d = 6 ( d = common difference )

we know,

=》

$nth \: \: term \: = a \: + (n - 1) \times d$

=》

$11th \: \: term = 5 + (11 - 1) \times 6$

=》

$11th \: \: term = 5 + (10 \times 6)$

=》

$11th \: \: term = 5 + 60$

=》

$11th \: \: term = 65$

nth term ( 11th term ) = 65

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its the 3rd answer!

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