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soldi70 [24.7K]
4 years ago
12

Frank and Mary are traveling to the beach which is 520 mi from their home if they travel at an average speed of 65 miles per hou

r and they don't make any stops how long will it take them to get there
Mathematics
1 answer:
Dvinal [7]4 years ago
4 0

Answer:

It will take 8 hours for Frank and Mary to get to the beach

Step-by-step explanation:

Step 1: Give expression for calculating

d=s×t

where;

d=distance traveled i miles

s=speed in miles per hour

t=time in hours

in our case;

d=520 miles

s=65 miles per hour

t=unknown

Step 2: Solve for the unknown, t

In order to solve for t, the expressions can be rearranged as follows ;

t=d/s

replacing values for d and s;

t=520/65

t=8 hours

It will take 8 hours for Frank and Mary to get to the beach

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Maisha has 3(x+2) hours reading a book 8x-5 hours sleeping and 2x hours playing.How many hours did she spend in all of these act
Mandarinka [93]
Add all the activities together

=3(x+2) + (8x-5) + 2x
multiply 3 by all in parentheses
=(3*x) + (3*2) + (8x-5) + 2x
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combine like terms
=(3x + 8x + 2x) + (6 - 5)
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ANSWER: 13x + 1

Hope this helps! :)
6 0
3 years ago
At a certain auto parts manufacturer, the Quality Control division has determined that one of the machines produces defective pa
11Alexandr11 [23.1K]

Answer:

Probability that fewer than 2 of these parts are defective is 0.604.

Step-by-step explanation:

We are given that at a certain auto parts manufacturer, the Quality Control division has determined that one of the machines produces defective parts 19% of the time.

A random sample of 7 parts produced by this machine is chosen.

The above situation can be represented through Binomial distribution;

P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....

where, n = number of trials (samples) taken = 7 parts

            r = number of success = fewer than 2

           p = probability of success which in our question is % of defective

                 parts produced by one of the machine, i.e; 19%

<em>LET X = Number of parts that are defective</em>

<u>So, it means X ~ Binom(n = 7, p = 0.19)</u>

Now, probability that fewer than 2 of these parts are defective is given by = P(X < 2)

    P(X < 2) = P(X = 0) + P(X = 1)

                  =  \binom{7}{0}\times 0.19^{0} \times (1-0.19)^{7-0}+ \binom{7}{1}\times 0.19^{1} \times (1-0.19)^{7-1}

                  =  1 \times 1 \times 0.81^{7} +7 \times 01.9^{1} \times 0.81^{6}

                  =  <u>0.604</u>

<em>Therefore, the probability that fewer than 2 of these parts are defective is 0.604.</em>

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irakobra [83]

Answering pretty sure yes

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