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3 years ago

What is the value of B?

Mathematics

1 answer:

DedPeter [7]3 years ago

4 0

Answer:

61°

Step-by-step explanation:

B = 180°-58°-61°

= 61°

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The diameters of bearings used in an aircraft landing gear assembly have a standard deviation of ???? = 0.0020 cm. A random samp

vovikov84 [41]

Answer:

(a) We conclude after testing that mean diameter is 8.2500 cm.

(b) P-value of test = 2 x 0.0005% = 1 x $10^{-5}$ .

(c) 95% confidence interval on the mean diameter = [8.2525 , 8.2545]

Step-by-step explanation:

We are given with the population standard deviation, $\sigma$ = 0.0020 cm

Sample Mean, Xbar = 8.2535 cm and Sample size, n = 15

(a) Let Null Hypothesis, $H_0$ : Mean Diameter, $\mu$ = 8.2500 cm

Alternate Hypothesis, $H_1$ : Mean Diameter, $\mu$ $\neq$ 8.2500 cm{Given two sided}

So, Test Statistics for testing this hypothesis is given by;

$\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ follows Standard Normal distribution

After putting each value, Test Statistics = $\frac{8.2535-8.2500}{\frac{0.0020}{\sqrt{15} } }$ = 6.778

Now we are given with the level of significance of 5% and at this level of significance our z score has a value of 1.96 as it is two sided alternative.

<em>Since our test statistics does not lie in the rejection region{value smaller than 1.96} as 6.778>1.96 so we have sufficient evidence to accept null hypothesis and conclude that the mean diameter is 8.2500 cm.</em>

(b) P-value is the exact % where test statistics lie.

For calculating P-value , our test statistics has a value of 6.778

So, P(Z > 6.778) = Since in the Z table the highest value for test statistics is given as 4.4172 and our test statistics has value higher than this so we conclude that P - value is smaller than 2 x 0.0005% { Here 2 is multiplied with the % value of 4.4172 because of two sided alternative hypothesis}

Hence P-value of test = 2 x 0.0005% = 1 x $10^{-5}$ .

Probability(-1.96 < N(0,1) < 1.96) = 0.95 { This indicates that at 5% level of significance our Z score will lie between area of -1.96 to 1.96.

P(-1.96 < $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96\frac{\sigma}{\sqrt{n} }$ < $Xbar - \mu$ < $1.96\frac{\sigma}{\sqrt{n} }$ ) = 0.95

P( $-Xbar-1.96\frac{\sigma}{\sqrt{n} }$ < $-\mu$ < $1.96\frac{\sigma}{\sqrt{n} }-Xbar$ ) = 0.95

P( $Xbar-1.96\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $Xbar+1.96\frac{\sigma}{\sqrt{n} }$ ) = 0.95

So, 95% confidence interval for $\mu$ = [ $Xbar-1.96\frac{\sigma}{\sqrt{n} }$ , $Xbar+1.96\frac{\sigma}{\sqrt{n} }$ ]

= [ $8.2535-1.96\frac{0.0020}{\sqrt{15} }$ , $8.2535+1.96\frac{0.0020}{\sqrt{15} }$ ]

= [8.2525 , 8.2545]

Here $\mu$ = mean diameter.

Therefore, 95% two-sided confidence interval on the mean diameter

= [8.2525 , 8.2545] .

6 0

3 years ago

Read 2 more answers

12. If ln(2x – 2) = 2t – 2, what is the value of x?

Katyanochek1 [597]

ln(2x – 2) = 2t – 2

2x-2=e^(2t-2)

x=e^(2t-2)/2+1

8 0

3 years ago

The annual rate of depreciation, x, on a car that was purchased for $9,000 and is worth $4,500 after 5 years can be found using

Zolol [24]

Answer:

Step-by-step explanation:

We know that exponential formula of depreciation

$A=P(1-x)^t$

where

P is the initial amount

x is the interest rate

A is the amount after t years

we are given

The annual rate of depreciation, x, on a car that was purchased for $9,000

so, P=9000

we can plug value it

$A=9000(1-x)^t$

we are given

when x=5 , A=4500

so, we can plug it and solve for x

$4500=9000(1-x)^5$

$\frac{9000\left(1-x\right)^5}{9000}=\frac{4500}{9000}$

$\left(1-x\right)^5=\frac{1}{2}$

$x=-\left(\frac{1}{2}\right)^{\frac{1}{5}}+1$

$x=0.12945$

so, interest rate is 13%

now, we can plug x

and we get

$A=9000(1-0.13)^t$

$A=9000(87)^t$

Graph:

6 0

3 years ago

Read 2 more answers

To the nearest tenth, find the perimeter of ABC with vertices A (-2,-2) B (0,5) and C (3,1)

Pavlova-9 [17]

the perimeter will then just be the sum of the distances of A, B and C, namely AB + BC + CA.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\qquadB(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad \qquadd = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}\\\\\\AB=\sqrt{[0-(-2)]^2+[5-(-2)]^2}\implies AB=\sqrt{(0+2)^2+(5+2)^2}\\\\\\AB=\sqrt{4+49}\implies \boxed{AB=\sqrt{53}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\B(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad C(\stackrel{x_1}{3}~,~\stackrel{y_1}{1})\\\\\\BC=\sqrt{(3-0)^2+(1-5)^2}\implies BC=\sqrt{3^2+(-4)^2}$

$\bf BC=\sqrt{9+16}\implies \boxed{BC=5}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\C(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\\\\\\CA=\sqrt{(-2-3)^2+(-2-1)^2}\implies CA=\sqrt{(-5)^2+(-3)^2}\\\\\\CA=\sqrt{25+9}\implies \boxed{CA=\sqrt{34}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\~\hfill \stackrel{AB+BC+CA}{\approx 18.11}~\hfill$

5 0

3 years ago

Pls help me asap :(

Andreas93 [3]

x^2-12x+(6)^2 is correct answer

8 0

3 years ago