How much Greater is the amount Jason paid than amount he charged

During his science experiment, Yuri measured the temperature of a water sample as 24 degrees Celsius. To determine the temperatu

mr Goodwill [35]

Answer:

75.2 degrees Fahrenheit

Step-by-step explanation:

F = StartFraction 9 Over 5 EndFraction C + 32

Substitute 24 for C.

Multiply the value of C by StartFraction 9 Over 5 EndFraction before adding 32.

24 Degrees Celsius is 75.2 Degrees Fahrenheit.

8 0

2 years ago

Read 2 more answers

Pedro owns a shrimp truck near the beach. He sells garlic shrimp for $8 a plate and spicy shrimp for $6 a plate. Each day Pedro

yKpoI14uk [10]

Answer:

C

Step-by-step explanation:

You can plug in each combination to see what would work

C is the only combination that satisfies all the constraints: it earns him more than $800 and is less than 120 plates.

7 0

2 years ago

In a library, the ratio of maths books to science books is 3 : 2 and the ratio of science books to history books is 1 : 3. If th

trapecia [35]

Answer:

60 maths books

Step-by-step explanation:

Let the following be:

Maths books = m
Science books = s
History books = h

And we have ratios and sum:

m/s = 3/2
s/h = 1/3
m+s+h = 220

From the ratios we get:

m = 3s/2
h= 3s

Considering the above in the sum of books:

m+s+h = 220
3s/2 + s + 3s = 220
3s + 2s + 6s = 2*220
11s = 440
s = 440/11
s = 40

Number of maths books:

m = 3s/2 = 3*40/2 = 60

6 0

2 years ago

Out of five men and five women, we form a committee consisting of four different people. Assuming that each committee of size fo

masha68 [24]

Answer:

1. 10/21

2. 11/42

3. 41/42

4. 2/15

Step-by-step explanation:

Probability defines the ratio of the number of favorable outcomes to the total number of outcomes.

Combination Operations

$5C2 = \frac{5! }{2! (5-2)!} = \frac{5*4*3! }{2! 3!} =\frac{5*4 }{2*1} =\frac{20}{2}= 10$

$10C4 = \frac{10! }{4! (10-4)!} = \frac{10*9*8*7*6! }{4! 6!} =\frac{10*9*8*7 }{4*3*2*1} =\frac{5040}{24}= 210$

$5C3 = \frac{5! }{3! (5-3)!} = \frac{5*4*3! }{3! 2!} =\frac{5*4 }{2*1} =\frac{20}{2}= 10$

$5C1 = \frac{5! }{1! (5-1)!} = \frac{5*4! }{1! 4!} =\frac{5 }{1} = 5$

$8C2 = \frac{8! }{2! (8-2)!} = \frac{8*7*6! }{2! 6!} =\frac{8*7 }{2*1} = 4*7= 28$

$5C4 = \frac{5! }{4! (5-4)!} = \frac{5*4! }{4! 1!} =\frac{5 }{1} = 5$

$5C0 = \frac{5! }{0! (5-0)!} = \frac{5! }{0! 5!} = 1$

1. Probability of forming a committee consisting of two men and two women

= $\frac{5C2 * 5C2}{10C4} =\frac{10*10}{210} =\frac{10}{21}$

2. Probability of forming a committee that has more women than men

(NOTE: Since the committee desired has only 4 members, forming a committee that has more women than men can only be possible if we have 4 women, 0 men or 3 women 1 man)

= $\frac{5C3 * 5C1 + 5C4*5C0}{10C4} = \frac{10 * 5 + 5*1 }{210} = \frac{50 + 5}{210}=\frac{55}{210} = \frac{11}{42}$

3. Probability of forming a committee that has at least one man

(NOTE: at least one man implies the committee must not have all women members )

= $1 - \frac{5C4}{10C4} = 1 - \frac{5}{210} = 1 - \frac{1}{42} = \frac{41}{42}$

4. Probability of forming a committee with both Alice and Bob are members of the committee (NOTE: If Alice and Bob are compulsory member of the committee, there are only two more people to join the committee from the remaining 8 people)

$=\frac{8C2}{10C4} =\frac{28}{210} =\frac{2}{15}$

it be like a coin toss ut a little more complicated. For every toss we have 4 outcomes from 210 possililities

6 0

2 years ago

A telephone survey uses a random digit dialing machine to call subjects. the random digit dialing machine is expected to reach a

yanalaym [24]

P( $r$ successes in $n$ trials) = ⁿ $C_{r}$ × $[P(Success)^{r}]$ × $[P(failure)]^{n-r}$

We have
$n=8$
$r=2$
$P(Success)=0.15$
$P(Failure)=0.85$

Substitute these values into the formula, we have
P( 2 success in 8 trials) = ⁸C₂ × (0.15)² × (0.85)⁶
P( 2 success in 8 trials) = 0.2376.... ≈ 0.24 = 24%

6 0

2 years ago