This is due in 30 mins pls help me

Please help me i rlly need help

Mumz [18]

Answer:

3

Step-by-step explanation:

Given a line with points; (2, 5) (3, 8).

1. Find the slope of the given line

The formula for finding the slope is:

$\frac{y_{2}-y_{1} }{x_{2} - x_{1}}$

Substitute in the values;

$x_{1} = 2\\y_{1} = 5\\x_{2} = 3\\y_{2} = 8$

$\frac{8-5}{3-2}$

simplify;

$\frac{3}{1}$

= 3

2. Find the slope of the parallel line;

Remember, when two lines are parallel, they run alongside each other, of infinitely long, but they never touch. Hence two parallel lines have the same slope. Therefore, the slope of a line that is parallel to the given one will also have the same slope as the given one, which is 3.

5 0

3 years ago

A manufacturer of detergent claims that the contents of boxes sold weigh on average at least 20 ounces. The distribution of weig

jenyasd209 [6]

Answer:

$z=\frac{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625$

The p value for this case would be given by:

$p_v =P(z$

Since the p value is higher than the significance level provided we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 20 ounces.

Step-by-step explanation:

Information provided

$\bar X=19.87$ represent the sample mean

$\sigma=0.4$ represent the population deviation

$n=25$ sample size

$\mu_o =68$ represent the value that we want to test

$\alpha=0.01$ represent the significance level

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true mean is at least 20 ounces, the system of hypothesis would be:

Null hypothesis: $\mu \geq 20$

Alternative hypothesis: $\mu < 20$

The statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$z=\frac{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625$

The p value for this case would be given by:

$p_v =P(z$

Since the p value is higher than the significance level provided we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 20 ounces.

7 0

3 years ago

Evaluate 4k2 + 3 when k = 5

lyudmila [28]

Answer:

103

Step-by-step explanation:

Evaluate 4k2 + 3

To evaluate 4k² + 3, when k = 5, which means when you sees k, put 5 in replacement

4k² + 3 = 4(5)² + 3

4(5×5) + 3

4(25) + 3

open the bracket

4 × 25 = 100

∴ 100 + 3 = 103

Please mark me brainliest

7 0

3 years ago

Read 2 more answers

A rectangle has a length 6 more than it's width if the width is decreased by 2 and the length decreased by 4 the resulting has a

Rashid [163]

Answer:

Length of original rectangle: 11 units.

$\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}$

$\text{Perimeter of new rectangle}=20$

Step-by-step explanation:

Let x represent width of the original rectangle.

We have been given that a rectangle has a length 6 more than it's width. S the length of the original rectangle would be $x+6$ .

We have been given that when the width is decreased by 2 and the length decreased by 4 the resulting has an area of 21 square units.

The width of new rectangle would be $x-2$ .

The length of new rectangle would be $x+6-4=x+2$ .

The area of new rectangle would be $(x+2)(x-2)$ .

Now we will equate area of new rectangle with 21 and solve for x as:

$(x+2)(x-2)=21$

Applying difference of squares, we will get:

$x^2-2^2=21$

$x^2-4=21$

$x^2-4+4=21+4$

$x^2=25$

Since width cannot be negative, so we will take positive square root of both sides.

$\sqrt{x^2}=\sqrt{25}$

$x=5$

Therefore, the width of original rectangle is 5 units.

Length of the original rectangle would be $x+6\Rightarrow x+5=11$ .

Therefore, the length of original rectangle is 11 units.

$\text{Area of original rectangle}=5\times 11$

$\text{Area of original rectangle}=55$

Therefore, area of the original rectangle is 55 square units.

Now we will find ratio of the original rectangle area to the new rectangle area as:

$\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}$

We know that perimeter of rectangle is two times the sum of length and width.

$\text{Perimeter of new rectangle}=2((x+2)+(x-2))$

$\text{Perimeter of new rectangle}=2((5+2)+(5-2))$

$\text{Perimeter of new rectangle}=2(7+3)$

$\text{Perimeter of new rectangle}=2(10)$

$\text{Perimeter of new rectangle}=20$

Therefore, the perimeter of the new rectangle is 20 units.

7 0

3 years ago

3(-4+x)<-33 I need to solve for x

Serjik [45]

Answer:

−7

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers