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3 years ago

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Dave and Mary depart from the bus station at 10 am traveling in opposite directions. Dave's bus is going 40 miles per hour and M

ary's is going 50 miles per hour. At what time will they be 270 miles apart?

2 answers:

loris [4]3 years ago

8 0

In 3 hours they will be 270 miles apart.

NemiM [27]3 years ago

8 0

Since Dave and Mary are going opposite directions every hour mary will be moving 50 miles away and Dave will be moving 40 miles away meaning they will move 90 miles away from each other every hour. So 270 is 90 times three so in three hours they will be 270 miles away so at 1 pm they will be 270 miles apart.

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Step-by-step explanation:

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3 years ago

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Find two numbers such that the sum of their squares is 100. The square of the larger number is equal to the sum of the product o

Shtirlitz [24]

Answer with Step-by-step explanation:

Let x and y are two numbers

According to question

$x^2+y^2=100$ ....(1)

Suppose x is larger than y

$x^2=x(4y+8x)$

$x=4y+8x$

$x-8x=4y$

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Substitute the value of y

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3 years ago

Evaluate 5!+2!. Thank you!

Ghella [55]

Answer:

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Step-by-step explanation:

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3 years ago

What is the slope of a line perpendicular to the line whose equation is x+ 3y = -15 Fully simplify your answer.

alukav5142 [94]

Answer:

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Step-by-step explanation:

We can find the slope of the given line in this two way:

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2) the second method is rewrite the equation in slope intercept form and find the therm that multiply x

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3 years ago

The speed with which utility companies can resolve problems is very important. GTC, the Georgetown Telephone Company, reports it

brilliants [131]

Answer:

(a) 11.25 and 1.68

(b) 0.1651

(c) 0.3903

(d) 0.6865

Step-by-step explanation:

We are given that GTC, the Georgetown Telephone Company, reports it can resolve customer problems the same day they are reported in 75% of the cases and suppose the 15 cases reported today are representative of all complaints.

This situation can be represented through Binomial distribution as;

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

where, n = number of trials (samples) taken = 15

r = number of success

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

which customer problems are resolved on the same day, i.e.;75%

So, here X ~ $Binom(n=15,p=0.75)$

(a) Expected number of problems to be resolved today = E(X)

E(X) = $\mu$ = n * p = 15 * 0.75 = 11.25

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(b) Probability that 10 of the problems can be resolved today = P(X = 10)

P(X = 10) = $\binom{15}{10}0.75^{10}(1-0.75)^{15-10}$

= $3003*0.75^{10} *0.25^{5}$ = 0.1651

(c) Probability that 10 or 11 of the problems can be resolved today is given by = P(X = 10) + P(X = 11)

= $\binom{15}{10}0.75^{10}(1-0.75)^{15-10}+\binom{15}{11}0.75^{11}(1-0.75)^{15-11}$

= $3003*0.75^{10} *0.25^{5} + 1365*0.75^{11} *0.25^{4}$ = 0.3903

(d) Probability that more than 10 of the problems can be resolved today is

given by = P(X > 10)

P(X > 10) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)

= $\binom{15}{11}0.75^{11}(1-0.75)^{15-11}+\binom{15}{12}0.75^{12}(1-0.75)^{15-12} + \binom{15}{13}0.75^{13}(1-0.75)^{15-13}+\binom{15}{14}0.75^{14}(1-0.75)^{15-14} + \binom{15}{15}0.75^{15}(1-0.75)^{15-15}$

= $1365*0.75^{11} *0.25^{4} + 455*0.75^{12} *0.25^{3}+105*0.75^{13} *0.25^{2} + 15*0.75^{14} *0.25^{1}+1*0.75^{15} *0.25^{0}$

= 0.6865

3 0

3 years ago