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3 years ago

A rectangle has a height of 4w^3 and a width of 5w^2-3w-4.

Mathematics

2 answers:

shutvik [7]3 years ago

8 0

Answer:

20w^5 - 12w^4 - 16w^3

Step-by-step explanation:

Area of rectangle= length× breadth(or height in this case) of the rectangle

Basically, you multiply both sides of the rectangle

So, area of rectangle= 4w^3(5w^2 -3w -4)

Expand.

Area of rectangle= 4w^3(5w^2) - 4w^3(3w) - 4w^3(4)

Then you open the brackets which will give you the answer 20w^5- 12w^4- 16w^3

Note that the numbers are all multiplied and for the powers of w, the powers are added when multiplied.

For example, y(y^2)= y^3

Thepotemich [5.8K]3 years ago

6 0

Answer:

Step-by-step explanation:

Area of the entire rectangle = height * width

= 4w³ *( 5w²-3w-4)

= 4w³ * 5w²- 4w³ *3w - 4w³ *4

=20w⁵ - 12w⁴ - 16w³

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4 years ago

The average American man consumes 9.8 grams of sodium each day. Suppose that the sodium consumption of American men is normally

Alex Ar [27]

Answer:

(a) The distribution of X is N (9.8, 0.8²).

(b) The probability that an American consumes between 8.8 and 9.9 grams of sodium per day is 0.4461.

Step-by-step explanation:

The random variable X is defined as the amount of sodium consumed.

The random variable X has an average value of, μ = 9.8 grams.

The standard deviation of X is, σ = 0.8 grams.

(a)

It is provided that the sodium consumption of American men is normally distributed.

The random variable X follows a normal distribution with parameters μ = 9.8 grams and σ = 0.8 grams.

Thus, the distribution of X is N (9.8, 0.8²).

(b)

If X ~ N (µ, σ²), then $Z=\frac{X-\mu}{\sigma}$ , is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z ~ N (0, 1).

To compute the probability of Normal distribution it is better to first convert the raw score (X) to z-scores.

Compute the probability that an American consumes between 8.8 and 9.9 grams of sodium per day as follows:

$P(8.8$

$=P(-1.25$

Thus, the probability that an American consumes between 8.8 and 9.9 grams of sodium per day is 0.4461.

(c)

The probability representing the middle 30% of American men consuming sodium between two weights is:

$P(x_{1}$

Compute the value of z as follows:

$P(-z$

The value of z for P (Z < z) = 0.65 is 0.39.

Compute the value of x₁ and x₂ as follows:

$-z=\frac{x_{1}-\mu}{\sigma}\\-0.39=\frac{x_{1}-9.8}{0.8}\\x_{1}=9.8-(0.39\times 0.8)\\x_{1}=9.488\\x_{1}\approx9.5$ $z=\frac{x_{2}-\mu}{\sigma}\\0.39=\frac{x_{1}-9.8}{0.8}\\x_{1}=9.8+(0.39\times 0.8)\\x_{1}=10.112\\x_{1}\approx10.1$

Thus, the middle 30% of American men consume between 9.5 grams to 10.1 grams of sodium.

4 0

3 years ago