F(x)= -1/3x+13 when value is f(-3)

How much wire is needed to build a rectangular pen 42.6 ft. long and 28.5 ft. wide

andreev551 [17]

Answer: 142.2 ft

Explanation:

Wire needed = Perimeter = 2(42.6 + 28.5) = 142.2 ft

4 0

4 years ago

Read 2 more answers

Shaun is 4 years older than her sister Charmaine. If the sum of their ages is 16, how old is Charmaine?

skad [1K]

Step-by-step explanation:

$let \: the \: age \: of \: charmaine \: be \: x$

$according \: to \: question$

$4 + x = 16$

$x = 16 - 4$

$12$

$so \: the \: age \: of \: shaun \: is \:$

$x + 4$

$12 + 4$

$16$

hope it helps if it helps don't forget to like and mark me

4 0

3 years ago

The population of a town is increasing by 2% each year. If the population today is 40,000, what

UkoKoshka [18]

Answer:

46,400

Step-by-step explanation:

present population =40,000

2% of 40,000 =800

for each year it increases by 800

in 8 years time =(8*800) +40,000

=46,400

5 0

3 years ago

Use vertical multiplication to find the product of:

Rom4ik [11]

Answer:

$x^6+x^4+4x^3-2x^2-x+3$

Step-by-step explanation:

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

First step multiply your terms in your first expression just to the 1 in the second expression like so:

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$x^3+2x+3$ Anything times 1 is that anything.

That is, $(x^3+2x+3) \cdot 1=x^3+2x+3$ .

Now we are going to take the top expression and multiply it to the -x in the second expression. $-x(x^3+2x+3)=-x^4-2x^2-3x$ . We are going to put this product right under our previous product.

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$x^3+2x+3$

$-x^4-2x^2-3x$

We still have one more multiplication but before we do that I'm going to put some 0 place holders in and get my like terms lined up for the later addition:

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$0x^4+x^3+0x^2+2x+3$

$-x^4+0x^3-2x^2-3x+0$

Now for the last multiplication, we are going to take the top expression and multiply it to x^3 giving us $x^3(x^3+2x+3)=x^6+2x^4+3x^3$ . (I'm going to put this product underneath our other 2 products):

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$0x^4+x^3+0x^2+2x+3$

$-x^4+0x^3-2x^2-3x+0$

$x^6+2x^4+3x^3$

I'm going to again insert some zero placeholders to help me line up my like terms for the addition.

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$0x^6+0x^4+x^3+0x^2+2x+3$

$0x^6-x^4+0x^3-2x^2-3x+0$

$x^6+2x^4+3x^3+0x^2+0x+0$

----------------------------------------------------Adding the three products!

$x^6+x^4+4x^3-2x^2-x+3$

8 0

4 years ago

Suppose that the weight of seedless watermelons is normally distributed with mean 6.2 kg. and standard deviation 1.5 kg. Let X b

Oksi-84 [34.3K]

Answer:

A.

X ~ N(6.2kg, 2.25kg²)

B. What is the median seedless watermelon weight?____ kg.

6.2 kg

C. What is the Z-score for a seedless watermelon weighing 8 kg?

1.2

D. What is the probability that a randomly selected watermelon will weigh more than 7 kg?

0.2981

E. What is the probability that a randomly selected seedless watermelon will weigh between 4 and 5 kg?

0.1411

F. The 80th percentile for the weight of seedless watermelons is _____ kg.

7.5 kg

Explanation:

A. X ~ N(___ , ____ )

The distribution of a random variable in a sample extracted from a population that follows a normal distribution is represented by the notation:

X ~ N(μ, σ²)

Where:

X is the random variable
N stands for normal distribution function
μ is the median of the population
σ² is the variance of the population

Here, you have:

μ = 6.2 kg

σ² = (1.5kg)² = 2.25 kg²

X ~ N(6.2kg, 2.25kg²)

B. What is the median seedless watermelon weight?____ kg.

The median of a random variable that follows a normal distribution is equal to the mean, thus it is 6.2 kg.

C. What is the Z-score for a seedless watermelon weighing 8 kg?

The z-score is the standardized value of the random variable. It measures how far away is the variable from the mean.

It is calcuated with the formula:

$Z-score(X=x)=\dfrac{x-\mu}{\sigma}$

Thus for X=8:

$Z(X=8)=\dfrac{8-6.2}{1.5}=1.2$

D. What is the probability that a randomly selected watermelon will weigh more than 7 kg?

You want P(X>7)

You must use the tables for the standardized normal distribution.

Find the corresponding Z-score for X = 7

$Z(X=7)=\dfrac{7-6.2}{1.5}\approx0.53$

You must use a table for the standardized normal distribution which gives the cumulative distribution or area under the curve of the standard normal distribution and find P(Z>0.53).

That is the area to the right of Z=0.53. The table shows 0.2981.

Thus, the probability that a randomly selected seedless watermelon will weigh more than 7 kg is 0.2981

E. What is the probability that a randomly selected seedless watermelon will weigh between 4 and 5 kg?

For this case you must find the Z-scores for X=4 and X=5 and then find the area under the curve of the standardized normal distribution between those two Z-scores.

Z(X=4) = (4 - 6.2)/1.5 ≈ -1.47

Z(X=5) = (5 - 6.2)/1.5 = -0.8

In the table the area to the right of Z = - 1.47 is 1 - 0.0708 = 0.9292

And the area to the right of Z = - 0.8 is 1 - 0.2119 = 0.7881

Thus, the area in between is the difference 0.9292 - 0.7881 = 0.1411.

F. The 80th percentile for the weight of seedless watermelons is _____ kg.

The 80th percentile is the weigh of the top 20% seedless watermelons: this is 80% of the weighs are below that weight.

You must find the Z-score for which the area under the curve is less than 0.80.

The area less than 0.80 is 1 less the area that is less than 0.20.

From the table, the Zscore that defines the area less than 0.20 is 0.845 (interpolating).

Thus, the 80th percentile is the X value that makes the Z-score greater than or equal to 0.845:

Z ≥ 0.845

(X - 6.2)/1.5 ≥ 0.845

X ≥ 0.845 × 1.5 + 6.2

X ≥ 7.4675

X ≥ 7.5 kg ← answer

7 0

3 years ago