All categories

3 years ago

The width of a rectangle is 5 n minus 1.5 feet and the length is 1.5 n plus 9 feet. Find the perimeter of the rectangle.

Mathematics

1 answer:

andre [41]3 years ago

5 0

Answer:

$Perimeter= 13n+15$

Step-by-step explanation:

We know that

Length(l) = 1.5n + 9 ft

Width(w) = 5n - 1.5 ft

The formula for the perimeter(P) of a rectangle is:

$P=2l+2b \\or \\P=2(l+b)$

So by substituting l and w we get :

$P= 2[(1.5n+9)+(5n-1.5)]$

$P= 2[1.5n+5n+9-1.5]$

$P= 2[6.5n+7.5]$

$P= 13n+15$

You might be interested in

HELP please.

Delvig [45]

Answer:

Q1: $\frac{3}{4}$

Q2: $\frac{83}{100}$

Step-by-step explanation:

Q1: 12 : 16 = 3 : 4 = $\frac{3}{4}$

Q2: 8.3 : 10 = $\frac{83}{100}$

3 0

3 years ago

Consider the probability that greater than 94 out of 153 people will not get the flu this winter. Assume the probability that a

Hatshy [7]

Answer:

0.8212 = 82.12% probability that greater than 94 out of 153 people will not get the flu this winter.

Step-by-step explanation:

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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .