Answer:
n = 5, n = -7/2
Step-by-step explanation:
2n + 5 = l
n - 4 = w
A = l × w = 15
(2n + 5)(n-4) = 15
2n^2 - 8n + 5n - 20 = 15
2n^2 -3n - 35 = 0
(2n + 7)(n - 5) = 0
2n + 7 = 0 or n - 5 = 0
n = -7/2 or n = 5
Using relations in a right triangle, it is found that the values of x and y are given by: x = 24, y = 46.4, given by option a.
<h3>What are the relations in a right triangle?</h3>
The relations in a right triangle are given as follows:
- The sine of an angle is given by the length of the opposite side to the angle divided by the length of the hypotenuse.
- The cosine of an angle is given by the length of the adjacent side to the angle divided by the length of the hypotenuse.
- The tangent of an angle is given by the length of the opposite side to the angle divided by the length of the adjacent side to the angle.
First, we start with the vertical line h that divides y, that is <u>opposite to an angle of 30º, with hypotenuse 34</u>, hence:
sin(30º) = h/34
0.5 = h/34
h = 17.
Then, h is opposite to an angle of 45º, while the hypotenuse is x, hence:


x = 24.
y is divided into two segments.
- The first is the adjacent to the angle of 30º, while the hypotenuse is 34.
- The second is adjacent to the angle of 45º, while the hypotenuse is 24.
Then:




Then, the value of y is given by:
.
More can be learned about relations in a right triangle at brainly.com/question/26396675
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1) f(x)=2x+6
f(2)=2(2)+6
=4+6
=10
2)f(x)=3x
f(a+1)=3(a+1)
=3a+3
3)f(x)=3x-1 and g(x)=5x+3
f(2)=3(2)-1
f(2)=6-1
=5
g(3)=5x+3
=5(3)+32
=15+3
=18
f(2)+f(3)=5+18
=23.
Answer:
a) ![[-0.134,0.034]](https://tex.z-dn.net/?f=%5B-0.134%2C0.034%5D)
b) We are uncertain
c) It will change significantly
Step-by-step explanation:
a) Since the variances are unknown, we use the t-test with 95% confidence interval, that is the significance level = 1-0.05 = 0.025.
Since we assume that the variances are equal, we use the pooled variance given as
,
where
.
The mean difference
.
The confidence interval is

![= -0.05\pm 1.995 \times 0.042 = -0.05 \pm 0.084 = [-0.134,0.034]](https://tex.z-dn.net/?f=%3D%20-0.05%5Cpm%201.995%20%5Ctimes%200.042%20%3D%20-0.05%20%5Cpm%200.084%20%3D%20%5B-0.134%2C0.034%5D)
b) With 95% confidence, we can say that it is possible that the gaskets from shift 2 are, on average, wider than the gaskets from shift 1, because the mean difference extends to the negative interval or that the gaskets from shift 1 are wider, because the confidence interval extends to the positive interval.
c) Increasing the sample sizes results in a smaller margin of error, which gives us a narrower confidence interval, thus giving us a good idea of what the true mean difference is.
How dos3es it want it in decimsal form there are a couple ways in decimal form at list i was tought