All categories

3 years ago

the ratio of boys to girls at the beach clean up was 7:8 if there are 42 boys how many girls are there

1 answer:

4 0

48 girls are there. Given that there is a ratio of boys to girls of 7:8. Then divide 42 boys by 7 boys which equals 6. After, multiply 6 by 8 girls equals 48 girls.

You might be interested in

Which equation can be used to solve the problem 180 is 40% of what number

Kruka [31]

Answer:

40/100 * x = 40

Solve for x. Hope this helps

6 0

4 years ago

Read 2 more answers

Plz i need help with this question

stellarik [79]

7x+9 is equivalent :)

4 0

3 years ago

Read 2 more answers

An aircraft manufacturer wants to determine the best selling price for a new airplane. The company estimates that the initial co

Blizzard [7]

Answer:

$(a)$ $C(x)=500+20x-5x^{\frac{3}{4}}+0.01x^2$

$p(x)=320-7.7p$

$R(x)=(320-7.7p)p=320p-7.7p^2$

$(b)$ $x=82 \text{planes}$

$(c)$ $p=\$30.91 M\;\; \text{per plane}$

$(d)$ maximum profit $=\$ 15.90M$

Step-by-step explanation:

Given that,

The company estimates that the initial cost of designing the aeroplane and setting up the factories in which to build it will be 500 million dollars.

The additional cost of manufacturing each plane can be modelled by the function.

$m(x)=20x-5x^{\frac{3}{4}}+0.01x^2$

$(a)$ Find the cost, demand (or price), and revenue functions.

$C(x)=500+20x-5x^{\frac{3}{4}}+0.01x^2$

$p(x)=320-7.7p$

$R(x)=(320-7.7p)p=320p-7.7p^2$

$(b)$ Find the production level that maximizes profit.

$f=R(x)-C(x)$

$\Rightarrow f=320p-7.7p^2-(500+20x-5x^{\frac{3}{4}}+0.01x^2)$

$\Rightarrow df=320dp-15.4pdp-20dx+5(\frac{3}{4} )x^{\frac{-1}{4} }dx-0.02xdx$

$x=320-7.7p$

$p=\frac{320-x}{7.7}$

$\frac{dp}{dx} = \frac{-1}{7.7}$

$\frac{df}{dx}=\frac{320}{-7.7} -\frac{15.4(320-x) }{7.7(\frac{-1}{7.7} )}-20+5\frac{3}{4} x^{\frac{-1}{4}} -0.02x=0$

$\Rightarrow -41.5584+83.1169-0.2597x-20+3.75x^{\frac{-1}{4} }-0.02x=0$

$\Rightarrow 21.5585+3.75x^{\frac{-1}{4} }-0.279x=0$

$\Rightarrow x=82 \text{planes}$

$(c)$ Find the associated selling price of the aircraft that maximizes profit.

$p=\frac{320-82}{7.7}$

$\Rightarrow p=\$30.91 M\;\; \text{per plane}$

$(d)$ Find the maximum profit.

Manufacturing cost of one plane is:

$m(1)=20-5+0.01$

$=\$15.01 M$

maximum profit $=\$(30.91-15.01)M$

$=\$15.90M$

3 0

3 years ago

Kevin has $1427 in a savings account. His brother, Harley has saved an amount with a 4 that is 1/10 the value of the 4 in Kevin'

Paul [167]

I'm guessing 400$????

3 0

3 years ago

A jar contains 12 red marbles numbered 1 to 12 and 6 blue marbles numbered 1 to 6. A marble is drawn at random from the jar. Fin

lora16 [44]

Answer:

$(a)P(Red)=\dfrac{2}{3}\\(b)P(Odd) =\dfrac{1}{2}\\(c)P(\text{Red or Odd Numbered})=\dfrac{5}{6}\\(d)P(\text{Blue or Even Numbered})=\dfrac{2}{3}$

Step-by-step explanation:

Number of Red Marbles{1,2,3,4,5,6,7,8,9,10,11,12},n(R)=12

Number of Blue Marbles{1,2,3,4,5,6},n(B)=6

Total Number of Marbles, n(S)=6+12=18

(a)Probability that the Marble is Red

$P(R)=\dfrac{n(R)}{n(S)} =\dfrac{12}{18} =\dfrac{2}{3}$

(b)Probability that the marble is odd-numbered.

Number of Odd-Numbered Balls, n(O)=9

$P(Odd)=\dfrac{n(O)}{n(S)} =\dfrac{9}{18} =\dfrac{1}{2}$

(c)Probability that the marble is red or odd-numbered.

n(Red)=12

n(Odd Numbered marbles)=9

n(Red and Odd Numbered Marbles)=6

$P(\text{Red or Odd Numbered})=P(Red)+P(Odd\:Numbered)-P(\text{Red and Odd Numbered)}\\=\dfrac{12}{18} +\dfrac{9}{18}-\dfrac{6}{18} =\dfrac{15}{18}\\P(\text{Red or Odd Numbered})=\dfrac{5}{6}$ (d)Probability that the marble is blue or even-numbered.

n(Blue)=6

n(Even Numbered marbles)=9

n(Blue and Even Numbered Marbles)=3

$P(\text{Blue or Even Numbered})=P(Blue)+P(Even\:Numbered)-P(\text{Blue and Even Numbered)}\\=\dfrac{6}{18} +\dfrac{9}{18}-\dfrac{3}{18} =\dfrac{12}{18}\\P(\text{Blue or Even Numbered})=\dfrac{2}{3}$

5 0

3 years ago