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3 years ago

Help asp please!!!!!!!!

Mathematics

2 answers:

ExtremeBDS [4]3 years ago

6 0

The answer is 87. If you want the work, say so in the comments. I'm just not writing it because I'm lazy but I will if you ask.

Fittoniya [83]3 years ago

3 0

87 is the correct answer. Just square 63 and 60 then add them. (comes out to 7569). Get the square root of that which then equals 87.

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Yuri writes this inequality:

Minchanka [31]

A. no
b. yes
c. no
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3 years ago

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The prime factorizations of 16 and 24 are shown below. Prime factorization of 16: 2, 2, 2, 2 Prime factorization of 24: 2, 2, 2,

postnew [5]

8 is your answer sir I just took the test got it right brainly plllzzz

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2 years ago

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GIVING BRAINLIEST TO WHOEVER GETS IT RIGHT

ahrayia [7]

Answer:

x=3 x = -1

Step-by-step explanation:

f(x) = x^2 − 2x − 3

Set the function equal to zero

0 = x^2 − 2x − 3

Factor

What two numbers multiply to -3 and add to -2

-3 *1 = -3

-3+1 = -2

0 = (x-3)(x+1)

Using the zero product property

x-3 =0 x+1 =0

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7 0

2 years ago

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find the height difference between the top of mt Everest 8848 metres and the Mariana Trench 10911 metresbelow sea level

zhuklara [117]

$Mount\ Everest\to+8,848m\\\\Mariana\ Trench\to-10,911m\\\\Difference:|-10,911-8,848m|=|-19,759|\\\\\center={\boxed{19,759\ (m)}}\leftarrow answer$

6 0

3 years ago

Consider the accompanying data on breaking load (kg/25mm width) for various fabrics in both an unabraded condition and an abrade

WITCHER [35]

Answer:

The critical value for a one-tailed test with 7 degrees of freedom and level of significance α=0.01 is tc=2.998.

The test statistic (t=1.729) is below the critical value and falls within the acceptance region, so it is failed to reject the null hypothesis.

At a significance level of 0.01, there is not enough evidence to support the claim that the true difference is significantly bigger than 0.

Step-by-step explanation:

We have a matched-pair t-test for the difference.

We have the null and alternative hypothesis written as:

$H_0: \mu_d=0\\\\H_a: \mu_d>0$

We have n=8 pairs of data. We calculate the difference as:

$d_1=U_1-A_1=36.4-28.5=7.9$

Then, with this procedure we get the sample for d:

$d=[7.9,\, 35,\, 5.5,\, 4.2,\, 6.7,\, -3.7,\, -0.9,\, 3.3]$

The sample mean and standard deviation are:

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{8}(7.9+35+5.5+. . .+3.3)\\\\\\M=\dfrac{58}{8}\\\\\\M=7.25\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{7}((7.9-7.25)^2+(35-7.25)^2+(5.5-7.25)^2+. . . +(3.3-7.25)^2)}\\\\\\s=\sqrt{\dfrac{985.08}{7}}\\\\\\s=\sqrt{140.73}=11.86\\\\\\$

Now, we can perform the one-tailed hypothesis test.

The significance level is 0.01.

The sample has a size n=8.

The sample mean is M=7.25.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.86.

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

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{11.86}{\sqrt{8}}=4.1931$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{7.25-0}{4.1931}=\dfrac{7.25}{4.1931}=1.729$

The degrees of freedom for this sample size are:

$df=n-1=8-1=7$

This test is a right-tailed test, with 7 degrees of freedom and t=1.729, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t>1.729)=0.0637$

As the P-value (0.0637) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.01, there is not enough evidence to support the claim that the true difference is significantly bigger than 0.

If we use the critical value approach, the critical value for a one-tailed test with 7 degrees of freedom and level of significance α=0.01 is tc=2.998. The test statistic is below the critical value and falls within the acceptance region, so it is failed to reject the null hypothesis.

8 0

3 years ago