Find the measure of angle m

What is the biggest whole number smaller than the perimeter of any triangle with a side of length 5.6 and a side of length 19.7?

snow_tiger [21]

Answer:

50

Step-by-step explanation:

The perimeter of a Triangle is the sum of it's 3 sides.

We are given two sides here which is side length 5.6 and a side of length 19.7

Let's us represent the third side as x

Therefore

x + 19.7 + 5.6 = Perimeter of the triangle

We would have this equation

But it is important to know that every side of a triangle must be less than the sum of the other two sides, hence

x < 19.7 + 5.6

x < 25.3

Adding 25.3 to both sides to make the left side equal to the perimeter

perimeter = x+25.3 < 25.3 + 25.3 = 50.6

Therefore, 50.6 is the smallest whole number that is larger than the perimeter of the triangle in the question above.

Therefore, the biggest whole number smaller than the perimeter of the above triangle is 50

6 0

3 years ago

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Can someone pls solve these for me i’ll give brainlist

Sedbober [7]

Answer

3. 14

4. 4

5. 8

Step-by-step explanation:

5 0

3 years ago

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Consider the following hypothesis test:

postnew [5]

Answer:

a. P-value = 0.039.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.

b. P-value = 0.013.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.

c. P-value = 0.130.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the population mean significantly differs from 100.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the population mean significantly differs from 100.

Then, the null and alternative hypothesis are:

$H_0: \mu=100\\\\H_a:\mu\neq 100$

The significance level is 0.05.

The sample has a size n=65.

The degrees of freedom for this sample size are:

$df=n-1=65-1=64$

a. The sample mean is M=103.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.5.

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

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{11.5}{\sqrt{65}}=1.4264$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{103-100}{1.4264}=\dfrac{3}{1.4264}=2.103$

This test is a two-tailed test, with 64 degrees of freedom and t=2.103, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>2.103)=0.039$

As the P-value (0.039) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.

b. The sample mean is M=96.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.

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

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{11}{\sqrt{65}}=1.3644$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{96.5-100}{1.3644}=\dfrac{-3.5}{1.3644}=-2.565$

This test is a two-tailed test, with 64 degrees of freedom and t=-2.565, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t$

As the P-value (0.013) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.

c. The sample mean is M=102.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=10.5.

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

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{10.5}{\sqrt{65}}=1.3024$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{102-100}{1.3024}=\dfrac{2}{1.3024}=1.536$

This test is a two-tailed test, with 64 degrees of freedom and t=1.536, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>1.536)=0.130$

As the P-value (0.13) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the population mean significantly differs from 100.

8 0

3 years ago

3 1/3 - 1 2/3 = ? Help me plzz i beg you i lost my notes :(

tamaranim1 [39]

You turn them into improper fractions by multiplying the bottom number to the whole number then adding the product to the top of the fraction, therefore
3 1/3 = 10/3, and 1 2/3 = 5/3

10 _ 5 = 5/3
3 3

8 0

3 years ago

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I will give many brainly

alex41 [277]

The 3rd one

Hope this helps

6 0

3 years ago