If three pencils cost total of 75cents, how much would six pencils

I need help with this question. can someone please help me

Ahat [919]

-2x + y =3
use 5 number and put it in above equation to find y, I will use -2,-1,0,1,2
x y
-2 -1
-1 1
0 3
1 5
2 7

Then draw the graph, find another point in the graph which intersect the slope,
lets say i found (-3,-3)which is on the slope
to prove it, i put (-3,-3) in the equation -2x + y = 3
and i got
3 = 3
Thus it is correct.

7 0

3 years ago

Bradley cut a square hole out of a block of wood in wood shop. If the block was cube-shaped with side lengths of 9 inches, and t

MatroZZZ [7]

To solve this problem, you must follow the proccedure below:

1. The block was cube-shaped with side lengths of 9 inches and to calculate its volume (V1), you must apply the following formula:

V1=s³

s is the side of the cube (s=9)

2. Therefore, you have:

V1=s³
V1=(9 inches)³
V1=729 inches³

3. The lengths of the sides of the hole is 3 inches. Therefore, you must calculate its volume (V2) by applying the formula for calculate the volume of a rectangular prism:

V2=LxWxH

L is the length (L=3 inches).
W is the width (W=3 inches).
H is the heigth (H=9 inches).

4. Therefore, you have:

V2=(3 inches)(3 inches)(9 inches)
V2=81 inches

5. The amount of wood that was left after the hole was cut out, is:

Vt=V1-V2
Vt=648 inches³

7 0

2 years ago

Which statement is true about the dependent and independent variables?

Hatshy [7]

Answer:

1. The independent variable goes on the x-axis and the dependent variable goes on the y-axis.

7 0

3 years ago

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.9 minutes and a standard deviation of 2.9

Eva8 [605]

Answer:

a) 0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) 0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes

c) 0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 8.9 minutes and a standard deviation of 2.9 minutes.

This means that $\mu = 8.9, \sigma = 2.9$

Sample of 37:

This means that $n = 37, s = \frac{2.9}{\sqrt{37}}$

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

320/37 = 8.64865

Sample mean below 8.64865, which is the p-value of Z when X = 8.64865. So

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

By the Central Limit Theorem

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

$Z = \frac{8.64865 - 8.9}{\frac{2.9}{\sqrt{37}}}$

$Z = -0.53$

$Z = -0.53$ has a p-value of 0.2981

0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

275/37 = 7.4324

Sample mean above 7.4324, which is 1 subtracted by the p-value of Z when X = 7.4324. So

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

$Z = \frac{7.4324 - 8.9}{\frac{2.9}{\sqrt{37}}}$

$Z = -3.08$

$Z = -3.08$ has a p-value of 0.001

1 - 0.001 = 0.999

0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Sample mean between 7.4324 minutes and 8.64865 minutes, which is the p-value of Z when X = 8.64865 subtracted by the p-value of Z when X = 7.4324. So

0.2981 - 0.0010 = 0.2971

0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

7 0

2 years ago

I am a number between 200 and 2006.I am odd and divisible by 3. What number am I?

Mashcka [7]

Answer: 201 is the correct answer

5 0

3 years ago