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maks197457 [2]
3 years ago
15

How many points are contained on a line?

Mathematics
1 answer:
inna [77]3 years ago
3 0

What kind of line? In order to answer this we have to know if its a graph line such as linear or if its a number line.

You might be interested in
If a square has the perimeter of 15.2 centimeters, what are 2 equations that can determine the perimeter of the square?
astraxan [27]

Answer:

1) P=4s

2) P=4\sqrt{A}

see the explanation

Step-by-step explanation:

we know that

The perimeter of the square is equal to

P=4s

where

s is the length side of the square

we have

P=15,2\ cm

substitute

15.2=4s

solve for s

s=15.2/4=3.8\ cm

The area of a square is equal to

A=s^2

substitute

A=3.8^2=14.44\ cm^2

we have that

s=\sqrt{A}

therefore

2 equations that can determine the perimeter of the square are

1) P=4s ----> P=4(3.8)=15.2\ cm

2) P=4\sqrt{A} ----> P=4\sqrt{14.44}=15.2\ cm

4 0
3 years ago
Factor completely 4x^5-20x^3
nikdorinn [45]

Answer:

<em>4x^3(x^2 - 5)</em>

Step-by-step explanation:

First, try to factor a common factor.

GCF of 4 and -20 is 4.

GCF of x^5 and x^3 is x^3.

Factor out 4x^3.

4x^5 - 20x^3 =

= 4x^3(x^2 - 5)

3 0
3 years ago
Only two full boxes and a box that was 75% full. Create a written expression
Alona [7]

Answer:

2( 100%) +75%

Step-by-step explanation:

a full box will literally be 100%, so two full boxes will be 2 × a full ie 100%, 'and' a box ie 75% full is addition '+'

5 0
2 years ago
What is the area of the trapezoid
anastassius [24]
 A= a+b/2xheight
8+14/2 x 12
=132in
8 0
3 years ago
Read 2 more answers
A random sample of n = 45 observations from a quantitative population produced a mean x = 2.5 and a standard deviation s = 0.26.
oee [108]

Answer:

P-value (t=2.58) = 0.0066.

Note: as we are using the sample standard deviation, a t-statistic is appropiate instead os a z-statistic.

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the population mean μ exceeds 2.4.

Then, the null and alternative hypothesis are:

H_0: \mu=2.4\\\\H_a:\mu> 2.4

The significance level is 0.05.

The sample has a size n=45.

The sample mean is M=2.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.26.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.26}{\sqrt{45}}=0.0388

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{2.5-2.4}{0.0388}=\dfrac{0.1}{0.0388}=2.58

The degrees of freedom for this sample size are:

df=n-1=45-1=44

This test is a right-tailed test, with 44 degrees of freedom and t=2.58, so the P-value for this test is calculated as (using a t-table):

P-value=P(t>2.5801)=0.0066

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

7 0
3 years ago
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