How to expand (5x−6)^2
2 answers:
You might be interested in
Answer:
x ≈ 3.5 cm
Step-by-step explanation:
Using the sine ratio in the right triangle
sin23° = =
=
( multiply both sides by 9 )
9 × sin23° = x , then
x ≈ 3.5 cm ( to 1 dec. place )
Answer:
<h3>present age of son = 10 </h3><h3>present age of father = 40</h3>Step-by-step explanation:
Let, present age of son be 'x'
present age of father be 'y'
y = 4x→ equation ( i )
After five years,
Son's age = x + 5
father's age = y + 5
According to Question,
Put the value of y from equation ( i )
Distribute 3 through the parentheses
Move variable to L.H.S and change it's sign
Similarly, Move constant to R.H.S. and change its sign
Collect like terms
Calculate the difference
Now, put the value of X in equation ( i ) in order to find the present age of father
plug the value of X
Calculate the product
Therefore,
Present age of son = 10
present age of father = 40
Hope this helps..
Best regards!!
Answer:
Critical value is t = 1.9901
Step-by-step explanation:
We are given the results of the sampling :
In-House Credit Card National Credit Card <u>Sample Size</u> : 32 50
<u>Mean Monthly Purchases </u>: $45.67 $39.87
<u>Standard Deviation </u>: $10.90 $12.47
Also, the managers wished to test whether there is a statistical difference in the mean monthly purchases by customers using the two types of credit cards, using a significance level α of 0.05.
<em>Firstly, we will specify our null and alternate hypothesis;</em>
Let = Mean Monthly Purchases of In-House Credit Card
= Mean Monthly purchases of National Credit Card
So, Null Hypothesis, :
= 0 {means that there is no difference in the mean monthly purchases by customers using the two types of credit cards}
Alternate Hypothesis, :
0 {means that there is statistical difference in the mean monthly purchases by customers using the two types of credit cards}
The test statistics that will be used here is <u>Two-sample t-test statistics</u>;
T.S. = ~
where, = Sample mean Purchases of In-House Credit Card = $45.67
= Sample mean Purchases of National Credit Card = $39.87
= pooled variance
= In-house credit card sample = 32
= National credit card sample = 50
So, degree of freedom of t-value here is (32 + 50 - 2) = 80
Now, at 0.05 significance level, t table gives critical value of t = 1.9901 at 80 degree of freedom.
<em>Therefore, the critical value assuming the population standard deviations are not known but that the populations are normally distributed with equal variances is t = 1.9901.</em>