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gregori [183]
3 years ago
8

In making a fruit punch for a party, you mixed 1⁄2 gallon of grape juice with 3⁄4 gallon of orange juice. How many gallons of pu

nch did you make?
Mathematics
2 answers:
Nikitich [7]3 years ago
6 0
3/4 + 1/2 = 1 1/4 . You give the same denominator in each fraction and add the numerator.
fiasKO [112]3 years ago
3 0

Answer:

I made 1.25 gallons of punch.

Step-by-step explanation:

As in making a fruit punch for a party, I mixed 1⁄2 gallon of grape juice with 3⁄4 gallon of orange juice, in order to know how many gallons of punch did I make we have to do the following calculation:

Grape juice: 1/2 = 0.5

Orange juice: 3/4 = 0.75

Total juice: grape juice + orange juice = 0.5 + 0.75

Total juice: 1.25 gallons of punch.

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Can someone help me find the answer and find out how to get the answer for X, Y, T, W, and Z
alexandr1967 [171]

would it be 90 ? because of the big angle

7 0
4 years ago
a company makes carboard cylindrical boxes with a diameter of 10 inches and a height of 8 inches.how many square inches of carbo
NeTakaya

Answer:

408.2sq inches

Step-by-step explanation:

Area of the cylindrical box = 2πr(r+h)

r is the radius  = diameter/2

r = 10/2 = 5in

h is the height = 8in

Substitute

Area of the cylindrical box  = 2(3.14)(5)(5+8)

Area of the cylindrical box = 2 * 3.14 * 5 * 13

Area of the cylindrical box  = 408.2sq inches

8 0
3 years ago
X and y are normal random variables with e(x) = 2, v(x) = 5, e(y) = 6, v(y) = 8 and cov(x,y)=2. determine the following: e(3x 2y
andriy [413]

The result for the given normal random variables are as follows;

a. E(3X + 2Y) = 18

b. V(3X + 2Y) = 77

c. P(3X + 2Y < 18) = 0.5

d. P(3X + 2Y < 28) = 0.8729

<h3>What is normal random variables?</h3>

Any normally distributed random variable having mean = 0 and standard deviation = 1 is referred to as a standard normal random variable. The letter Z will always be used to represent it.

Now, according to the question;

The given normal random variables are;

E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8.

Part a.

Consider E(3X + 2Y)

\begin{aligned}E(3 X+2 Y) &=3 E(X)+2 E(Y) \\&=(3) (2)+(2)(6 )\\&=18\end{aligned}

Part b.

Consider V(3X + 2Y)

\begin{aligned}V(3 X+2 Y) &=3^{2} V(X)+2^{2} V(Y) \\&=(9)(5)+(4)(8) \\&=77\end{aligned}

Part c.

Consider P(3X + 2Y < 18)

A normal random variable is also linear combination of two independent normal random variables.

3 X+2 Y \sim N(18,77)

Thus,

P(3 X+2 Y < 18)=0.5

Part d.

Consider P(3X + 2Y < 28)

Z=\frac{(3 X+2 Y-18)}{\sqrt{77}}

\begin{aligned} P(3X + 2Y < 28)&=P\left(\frac{3 X+2 Y-18}{\sqrt{77}} < \frac{28-18}{\sqrt{77}}\right) \\&=P(Z < 1.14) \\&=0.8729\end{aligned}

Therefore, the values for the given normal random variables are found.

To know more about the normal random variables, here

brainly.com/question/23836881

#SPJ4

The correct question is-

X and Y are independent, normal random variables with E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8. Determine the following:

a. E(3X + 2Y)

b. V(3X + 2Y)

c. P(3X + 2Y < 18)

d. P(3X + 2Y < 28)

8 0
2 years ago
Write each expression in standard form .<br> <br> -3(1-8m-2n)<br> 5-7(-4q+5)<br> -(2h-9)-4h
oksano4ka [1.4K]
-3(-8m-2n+1^5)-6h+28q-26
7 0
3 years ago
Give the values of Xmin, Xmax, Ymin, and Ymax for the screen, given the values for Xscl and Yscl.
joja [24]

Answer: Option C.

Step-by-step explanation:

Xmin, Xmax, Ymin, and Ymax represent the minimum/maximum values that X and Y can take.

We know that the scale in Y and X is 10.

This means that each mark in each axis represents 10 units.

Now, let's go to the vertical axis (y-axis)

Counting from the (0, 0) we can count:

4 marks up

4 marks down.

So if each mark represents 10 units, 4 marks are 40 units.

Then the range for Y is {-40, 40}

Now for the horizontal axis.

We can count 3 marks on each side, and with the same reasoning as before, we can find that the range for X is: {-30, 30}

Then we have that:

Xmin = -30

Xmax = 30

Ymin = -40

Ymax = 40

Now, as a point is defined as (X, Y)

This region can be written as:

{Xmin, Xmax} by {Ymin, Ymax}

or, replacing the values:

{-30, 30} by {-40, 40)

So the correct option is C.

4 0
3 years ago
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