Graph <br> −7y+8=21x−6<br> .

JulsSmile [24]

Resolving horizontally . Make forces to the right positive
resultant force = 70cos30-60cos60 = 60.62-60 = 0.62 lbs

resolve vertically
resultant force = 120 sin60 - 70 sin30 = 103.92 - 35 = 68.92 lbs in downward direction

magnitude of the ressultant forsse = sqrt(68.92^2 + 0.62^2) = 68.92 lbs

direction of the force = tan-1 (0.62 / 68.82) = 0.52 degrees east of south or
a bearing of
90 - 0.52 = 89.48 degrees

6 0

3 years ago

Two processes are used to produce forgings used in an aircraft wing assembly. Of 200 forgings selected from process 1, 10 do not

tangare [24]

Answer:

a.

$\bar p_1=0.05\\\bar p_2=0.067$

b-Check illustration below

c.(-0.0517,0.0177

Step-by-step explanation:

a.let $p_1 \& p_2$ denote processes 1 & 2.

For $p_1$ : T1=10,n1=200

For $p_2$ :T2=20,n2=300

Therefore

$\bar p_1=\frac{t_1}{N_1}=\frac{10}{200}=0.05\\\bar p_2=\frac{t_2}{N_2}=\frac{20}{300}=0.067$

b. To test for hypothesis:-

i.

$H_0:p_1=p_2\\H_A=p_1\neq p_2\\\alpha=0.05$

ii.For a two sample Proportion test

$Z=\frac{\bar p_1-\bar p_2}{\sqrt(\bar p(1-\bar p)(\frac{1}{n_1}+\frac{1}{n_2})}\\$

iii. for $\frac{\alpha}{2}=(-1.96,+1.96)$ (0.5 alpha IS 0.025),

reject $H_o$ if $|Z|>1.96$

iv. Do not reject $H_o$ . The noncomforting proportions are not significantly different as calculated below:

$z=\frac{0.050-0.067}{\sqrt {(0.06\times0.94)\times \frac{1}{500}}}$

z=-0.78

c. $(1-\alpha).100\%$ for the p1-p2 is given as:

$(\bar p_1-\bar p_2)\pm Z_0_._5_\alpha \times \sqrt \frac{ \bar p_1(1-\bar p_1)}{n_1}+\frac{\bar p_2(1-\bar p_2)}{n_2}\\\\=(0.05-0.067)\pm 1.645 \times \sqrt \ \frac{0.05+0.95}{200}+\frac{0.067+0.933}{300}\\$

=(-0.0517,+0.0177)

*CI contains o, which implies that proportions are NOT significantly different.

4 0

2 years ago

:

Alinara [238K]

There was 4.35% error in Diego's measurements.

Step-by-step explanation:

Given,

Length measured by Diego = Approx value = 22 cm

Actual length = Exact value = 23 cm

Percent error = $\frac{|approx-exact|}{exact}*100$

$Percent\ error = \frac{|22-23|}{23}*100\\\\Percent\ error = \frac{|-1|}{23}*100\\\\Percent\ error = \frac{1}{23}*100\\\\Percent\ error = \frac{100}{23}\\\\Percent\ error = 4.347\%$

Rounding off to nearest hundredth

Percent error = 4.35%

There was 4.35% error in Diego's measurements.

Keywords: percentage, subtraction

Learn more about percentages at:

#LearnwithBrainly

5 0

2 years ago

-x^5 divided by the square root of 12x-9

Vladimir79 [104]

-8.66667266 is the answer to you question.

7 0

2 years ago

On a given day, machine A has a 10% chance of malfunctioning and machine B has a 7% chance of the same. Given that at least one

valentinak56 [21]

I will give an instance to answer this. Suppose that we have 1,000 days. And it was given above that machine A has 10 % chances to malfunctioned and B has a 7%. So machine A has 100 days that malfunctioned while machine B has 70 days. And 7 days that both machines malfunctioned (10%x7%). Next is to add 100 days from machine A and 70 days from machine B and subtract with the 7 days where both of them malfunctioned. So the result is 163 days out of 1000 that we expect an incidence of failure. So the chance of machine B to malfunctioned is 70 days. 70/163 is 42.93%

4 0

2 years ago

Graph ​ −7y+8=21x−6 ​.

Graph −7y+8=21x−6 .