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Mandarinka [93]
3 years ago
11

Elena rode her bike 2 miles in 10 minutes. She rode at a constant speed. Complete the table to show the time it took her to trav

el different distances at this speed
Mathematics
2 answers:
harkovskaia [24]3 years ago
8 0

1 mile per 5 minutes

2:10

3:15

4:20 (Nice)

5:25

6:30

7:35

8:40

9:45

10:50

so on and so forth.

statuscvo [17]3 years ago
8 0

Answer:

what is question i do not get it

here is it

Step-by-step explanation:

Since you didn't give a picture of the table, I'll just show how you create a ratio table based off of the fact that Lin can ride her bike for 2 miles in 8 minutes

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HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Juli2301 [7.4K]

Answer:

4/8

Step-by-step explanation:

8 is how many cookies in total and 4 is how many that needs to be distributed.

7 0
3 years ago
You are dealt two cards successively without replacement from a standard deck of 52 playing cards. find the probability that the
Alenkasestr [34]
We have four two cards in 52 cards and four ten cards in the same playing cards 
to get our first four cards we choose one from the whole set of 52 cards
but to get the next card with ten we choose them out of 51 because we took the first card with value two and we have not replaced it
this gives us
4/52*4/51=0.006
7 0
3 years ago
What is the length of AC
Alex777 [14]

Answer:

13

Step-by-step explanation:

count A up to C

6 0
3 years ago
The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 mont
BartSMP [9]

Answer:

See steps below

Step-by-step explanation:

Let X be the random variable that measures the lifespan of a bulb.

If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is

\bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)

and its cumulative distribution function (CDF) is

\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}

• What is probability that the red bulb will need to be replaced at the first inspection?

The probability that the bulb fails the first year is

\bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347

• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?

Let A and B be the events,

A = “The bulb will last at least 24 months”

B = “The bulb will last at least 18 months”

We want to find P(A | B).

By definition P(A | B) = P(A∩B)P(B)

but B⊂A, so  A∩B = B and  

\bf P(A | B) = P(B)P(B) = (P(B))^2

We have  

\bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237

hence,

\bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313

• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent

If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347

Now the probability that exactly k bulbs need replacement is

\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}

<em>Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection. </em>

This means that,

<em>Probability that at least one of them needs replacement at the first inspection =  </em>

\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021

5 0
3 years ago
PLEASE HELP!!! Y=f(x)=(4)^x find f(x) when x=1/2
UNO [17]
\left[Y \right] = \left[ f(x)\right][Y]=[f(x)] .
I hope helping with u this answer
5 0
3 years ago
Read 2 more answers
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