The slope of the line is -3/4, so the answer is B
Answer:
x = 3
Step-by-step explanation:
Here, we want to find the value of x
By mathematical convention;
NM = NL + LM
Now, substitute individual values
3x + 13 = (6x-5) + (2x + 3)
3x + 13 = 6x-5 + 2x + 3
3x + 13 = 8x -2
Collect like terms
8x -3x = 13 + 2
5x = 15
x = 15/5
x = 3
The answer is: [A]: 45x² + 81x + 36 .
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Explanation: Use "FOIL" technique: (First, Outer, Inner, and Last Terms);
then, combine the "like terms", to simplify.
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(5x+4)(9x+9) = 45x² + 45x + 36x + 35 = 45x² + 81x + 36 .
(which is answer choice: "A".).
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Answer:
a) 3.128
b) Yes, it is an outerlier
Step-by-step explanation:
The standardized z-score for a particular sample can be determined via the following expression:
z_i = {x_i -\bar x}/{s}
Where;
\bar x = sample means
s = sample standard deviation
Given data:
the mean shipment thickness (\bar x) = 0.2731 mm
With the standardized deviation (s) = 0.000959 mm
The standardized z-score for a certain shipment with a diameter x_i= 0.2761 mm can be determined via the following previous expression
z_i = {x_i -\bar x}/{s}
z_i = {0.2761-0.2731}/{ 0.000959}
z_i = 3.128
b)
From the standardized z-score
If [z_i < 2]; it typically implies that the data is unusual
If [z_i > 2]; it means that the data value is an outerlier
However, since our z_i > 3 (I.e it is 3.128), we conclude that it is an outerlier.
Answer:
There is enough evidence to support the claim that the population mean is greater than 100
Step-by-step explanation:
<u>Step 1</u>: We state the hypothesis and identify the claim
and 
(claim)
<u>Step 2</u>: Calculate the test value.
<u>Step 3</u>: Find the P-value. The p-value obtained from a calculator is using d.f=39 and test-value 1.126 is 0.134
<u>Step 4</u>: We fail to reject the null hypothesis since P-value is greater that the alpha level. (0.134>0.05).
<u>Step 5</u>: There is enough evidence to support the claim that the population mean is greater than 100.
<u>Alternatively</u>: We could also calculate the critical value to obtain +1.685 for 
and d.f=39 and compare to the test-value:
The critical value (1.685>1.126) falls in the non-rejection region. See attachment.
NB: The t- distribution must be used because the population standard deviation is not known.
