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forsale [732]
3 years ago
12

The cost of 4 pounds of bananas is $3.52. What is the constant of proportionality that relates the cost of dollars, y, to the nu

mber of pounds of bananas, x?

Mathematics
1 answer:
Sladkaya [172]3 years ago
6 0
.88
3.52 divided by 4
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What's the easiest way to answer how I know the answer pls?​
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3 years ago
You can buy 3 pounds of strawberries for 3.99 at Kroger knowing this what is the unit rate per pound of strawberries and how muc
Varvara68 [4.7K]

Answer:

  • 1.33 per pound
  • 6.65 for 5 pounds

Step-by-step explanation:

To find cost per pound, divide cost by pounds.

  3.99/(3 lb) = 1.33/lb . . . . cost per pound

The cost of 5 pounds will be found by multiplying that by 5:

  (1.33/lb)(5 lb) = 6.65 . . . cost of 5 pounds

7 0
2 years ago
28 is equal to _________ hundredths.
andriy [413]

Answer: 28,09

Step-by-step explanation:

7 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
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