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2 years ago

How many grapes? Help me please! Thanks

Mathematics

1 answer:

Lena [83]2 years ago

4 0

Using proportions, considering the given ratio, it is found that 9090 grapes were there.

<h3>What is a proportion?</h3>

A proportion is a fraction of a total amount, and the measures are related using a rule of three.

The ratio of bananas to grapes of 3:101 means that out of 104 fruits, 3 are bananas and 101 are grapes. Hence the<em> rule of three</em> is given as follows:

3 bananas - 101 grapes

270 bananas - n grapes.

Applying cross multiplication:

3n = 101 x 270

n = 101 x 270/3

n = 9090.

9090 grapes were there.

More can be learned about proportions at brainly.com/question/24372153

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Can anyone explain how this problem is done ?

Svetach [21]

Step-by-step explanation:

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3 years ago

The height of a triangular window is 3 feet less than its base. The area the window is 20 ft.? find the dimensions of the window

Ivenika [448]

Let us assume base of the window = b feet.

Height is 3 feet less than base, that is = (b-3).

Given area of the triangle = 20 feet square.

We know formula for area of a triangle:

Area = $\frac{1}{2}$ × Base × Height

Plugging values in formula, we get

20 = $\frac{1}{2} b \times (b-3)$

$\mathrm{Multiply\:both\:sides\:by\:}2$

$40=b\left(b-3\right)$

$40=b^2-3b$

$b^2-3b-40=0$

Factoring quadratic, we get

$\left(b+5\right)\left(b-8\right)$ =0

b+5=0

b=-5

b-8=0

b=8.

For base we can take only positive value.

<h3>Therefore, base of the triangle = 8 feet and height = 8 -3 = 5 feet.</h3>

6 0

2 years ago

A car dealership sells 0, 1, or 2 luxury cars on any day. When selling a car, the dealer also tries to persuade the customer to

melisa1 [442]

Answer:

Mean = 1.42

Variance = 0.58

Step-by-step explanation:

Given: X denote the number of luxury cars sold in a given day, and Y denote the number of extended warranties sold.

Also, joint probability function of X and Y are given.

To find:

mean and variance of X

Solution:

From the given joint probability function of X and Y,

$P(X=0)=\frac{1}{6}\\P(X=1)=\frac{1}{12}+\frac{1}{6}=\frac{1+2}{12}=\frac{3}{12}\\P(X=2)=\frac{1}{12}+\frac{1}{3}+\frac{1}{6}=\frac{1+4+2}{12}=\frac{7}{12}$

Mean of X:

$E(X)=\sum XP(X)\\=0\left ( \frac{1}{6} \right )+1\left ( \frac{3}{12} \right )+2\left ( \frac{7}{12} \right )\\=0+\frac{3}{12}+\frac{14}{12}\\=\frac{17}{12}=1.42$

Variance of X:

$E(X^2)=\sum X^2P(X)\\=0^2\left ( \frac{1}{6} \right )+1^2\left ( \frac{3}{12} \right )+2^2\left ( \frac{7}{12} \right )\\=0+\frac{3}{12}+\frac{28}{12}\\=\frac{31}{12}$

$var(X)=E\left [ X^2 \right ]-\left ( E\left [ X \right ] \right )^2\\=\frac{31}{12}-\left ( \frac{17}{12} \right )^2\\=\frac{31}{12}-\frac{289}{144}\\=\frac{372-289}{144}\\=\frac{83}{144}\\=0.58$