Linda should use ...
15 gallons of 80% juice
5 gallons of 60% juice

_____
Let g represent the number of gallons of 80% juice needed. Then (20-g) is the number of gallons of 60% juice Linda will use. The amount of juice in the mixture can be written as
0.80g +0.60(20 -g) = 0.75×20
0.20g = 20(0.75 -0.60) = 20×0.15
g = 20×0.15/0.20 = 15
The amount of 80% juice needed is 15 gallons, so 5 gallons of 60% juice are needed.

5 0

3 years ago

Catherine invested a principal of $1,650 in her bank account with interest rate 3.1 % How much interest did she earn in 14 years

never [62]

Answer:

$693

Step-by-step explanation:

Catherine invested a principal of $1,650 in her bank account with;

interest rate of 3.1%

How much interest did she earn in 14 years?

To find the amount accumulated in the 14 years, we use the formula:

A = P(1 + rt)

Where A is the amount accumulated, P is the principal, r is the interest rate and t is the time.

A = $1650(1 + $\frac{3}{100}$ (14))

A = $1650 + $693 = $2343

Interest = Amount (A) - Principal (P) = $2343 - $1650 = $693

7 0

3 years ago

Read 2 more answers

The Town of Hertfordshire clerk knows that 23% of dogs in the town have completed emotional support training. Hertfordshire plan

Nataly_w [17]

Answer:

95.64% probability that under 30% of the dogs are emotional support trained

Step-by-step explanation:

For each dog, there are only two possible outcomes. Either they have completed emotional support training, or they have not. So we use the binomial probability distribution to solve this problem.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .