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2 years ago

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There are two types of ice cream cones - strawberry and peach. You and 24 of your closest friends are going to buy ice cream con

es. If 15 people buy strawberry cones and 20 people buy peach cones, how many people bought both strawberry and peach cones?

1 answer:

Zolol [24]2 years ago

6 0

Answer:

10 people bought both strawberry and peach cones

Step-by-step explanation:

Given that there are two types of ice cream cones - strawberry and peach. You and 24 of your closest friends are going to buy ice cream cones.

Assuming that each one had atleast one ice cream we get

$n(AUB) =24+1 =25$

A - no of people bought straw berry

B - no of people bought peach cones

n(A) = 15 and n(B) =20

Hence we have

$n(AUB)=n(A)+n(B)-n(AB)\\25= 15+20-n(AB)\\n(AB) = 10$

10 people bought both strawberry and peach cones

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Evgen [1.6K]

Answer:

x = -4, y = -6

Step-by-step explanation:

-x+2y = -8

3x-y = -6

x = 2y+8

3(2y+8) - y = -6

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3 years ago

Read 2 more answers

Two brothers are sharing a certain sum of money in the ratio 2/3 : 3/5, if the largest share was Gh¢500.00. How much was shared

lina2011 [118]

Given:

Two brothers are sharing a certain sum of money in the ratio $\dfrac{2}{3}:\dfrac{3}{5}$ .

The largest share was Gh¢500.00.

To find:

The total amount shared between them.

Solution:

It is given that, two brothers are sharing a certain sum of money in the ratio $\dfrac{2}{3}:\dfrac{3}{5}$ .

First we need to make the denominators common.

$Ratio=\dfrac{2\times 5}{3\times 5}:\dfrac{3\times 3}{5\times 3}$

$Ratio=\dfrac{10}{15}:\dfrac{9}{15}$

$Ratio=10:9$

So, the two brothers are sharing a certain sum of money in the ratio 10:9.

Let the shares of two brothers are 10x and 9x respectively. So, the larger share is 10x.

It is given that the largest share was Gh¢500.00.

$10x=500$

$x=\dfrac{500}{10}$

$x=50$

Now, the total amount shared between them is:

$Total=10x+9x$

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$Total=19(50)$

$Total=950$

Therefore, the total amount shared between them is Gh¢950.00.

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2 years ago

Find the sum of 1 3/7 plus 10 6/7

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Answer:

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Step-by-step explanation:

3/7 + 6/7 = 9/7

Change 9/7 to a mixed number = 1 2/7

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3 years ago

Use cylindrical coordinates to evaluate the triple integral ∭ where E is the solid bounded by the circular paraboloid z = 9 - 16

4vir4ik [10]

Answer:

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

Step-by-step explanation:

The Cylindrical coordinates are:

x = rcosθ, y = rsinθ and z = z

From the question, on the xy-plane;

$9 -16 (x^2 + y^2) = 0 \\ \\ 16 (x^2 + y^2) = 9 \\ \\ x^2+y^2 = \dfrac{9}{16}$

$x^2+y^2 = (\dfrac{3}{4})^2$

where:

0 ≤ r ≤ $\dfrac{3}{4}$ and 0 ≤ θ ≤ 2π

∴

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} \int ^{9-16r^2}_{0} \ r \times rdzdrd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 z|^{z= 9-16r^2}_{z=0} \ \ \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 ( 9-16r^2}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} ( 9r^2-16r^4}) \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( \dfrac{9r^3}{3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3r^3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0} \ drd \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} ( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) d \theta$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) \theta |^{2 \pi}_{0}$

$\iiint_E E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{64}}-\dfrac{243}{320}})2 \pi$

$\iiint_E E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{160}})2 \pi$

$\mathbf{\iiint_E E \sqrt{x^2+y^2} \ dV =\dfrac{81 \ \pi}{80}}$

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2 years ago