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2 years ago

3/10 of a bag of jelly beans 2/10 and of the bag were orange. What fraction of the bag was red and orange

Mathematics

1 answer:

Elden [556K]2 years ago

5 0

Answer:

1/2 of the bag would be red and orange.

Step-by-step explanation:

if 3/10 of the bag was red

and 2/10 was orange, then all you have to do is the add the number of jelly beans that are red and orange,

3/10 + 2/10 is 5/10

simplify and that will get you to 1/2

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Jamal filled his gas tank and then use 7/16 of the tank for traveling to visit his grandfather. He then used 1/3 of the remainin

alisha [4.7K]

Answer:

The fraction of the tank is filled with gasoline is $\frac{3}{8}$ .

Step-by-step explanation:

Consider the provided information.

Jamal filled his gas tank and then use 7/16 of the tank for traveling to visit his grandfather.

After his first traveling the amount of gas left can be calculated as:

$1-\frac{7}{16}$

$\frac{16-7}{16}$

$\frac{9}{16}$

He then used 1/3 of the remaining gas in the tank to run around town.

That means he left with 2/3 of gas.

$\frac{9}{16}\times \frac{2}{3}$

$\frac{3}{8}$

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3 years ago

two years local internet service costs $685, including the installation fee of $85. what is the monthly fee?

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2 years ago

A manufacturer knows that on average 20% of the electric toasters produced require repairs within 1 year after they are sold. Wh

Effectus [21]

Answer:

(a) The value of x is 5.

(b) The value of y is 15.

Step-by-step explanation:

Let the random variable X represent the number of electric toasters produced that require repairs within 1 year.

And the let the random variable Y represent the number of electric toasters produced that does not require repairs within 1 year.

The probability of the random variables are:

P (X) = 0.20

P (Y) = 1 - P (X) = 1 - 0.20 = 0.80

The event that a randomly selected electric toaster requires repair is independent of the other electric toasters.

A random sample of n = 20 toasters are selected.

The random variable X and Y thus, follows binomial distribution.

The probability mass function of X and Y are:

$P(X=x)={20\choose x}(0.20)^{x}(1-0.20)^{20-x}$

$P(Y=y)={20\choose y}(0.20)^{20-y}(1-0.20)^{y}$

(a)

Compute the value of x such that P (X ≥ x) < 0.50:

$P (X \geq x) < 0.50\\\\1-P(X\leq x-1)$

Use the Binomial table for n = 20 and p = 0.20.

$0.411=\sum\limits^{3}_{x=0}[b(x,20,0.20)]$

The least value of x that satisfies the inequality P (X ≥ x) < 0.50 is:

x - 1 = 4

x = 5

Thus, the value of x is 5.

(b)

Compute the value of y such that P (Y ≥ y) > 0.80:

$P (Y \geq y) >0.80\\\\P(Y\leq 20-y)>0.80\\\\P(Y\leq 20-y)>0.80\\\\\sum\limits^{20-y}_{y=0}[{20\choose y}(0.20)^{20-y}(1-0.20)^{y}]>0.80$

Use the Binomial table for n = 20 and p = 0.20.

$0.630=\sum\limits^{4}_{y=0}[b(y,20,0.20)]$

The least value of y that satisfies the inequality P (Y ≥ y) > 0.80 is:

20 - y = 5

y = 15

Thus, the value of y is 15.

3 0

3 years ago