Classify a triangle whose side lengths are 8, 16, and 18

Solve a÷b(×)-c=0 for x<br>Help me please :)

adell [148]

A b x−c=0 Step 1: Multiply both sides by b. ax−bc=0 Step 2: Add bc to both sides. ax−bc+bc=0+bc ax=bc Step 3: Divide both sides by x. ax x = bc x a= bc x Answer: a= bc x

3 0

3 years ago

18 minutes left hwhsusbe

Pepsi [2]

Answer:

so its D

Step-by-step explanation:

a is wrong becuase it does have 6

b is wrong becuase you make into 6 triangles

c is wrong becuase it does add up to 180 degrees

6 0

3 years ago

A machine that produces a major part for an airplane engine is monitored closely. In the past, 10% of the parts produced would b

nata0808 [166]

Answer:

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

For this problem, we have that:

$p = 0.1$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is

We need a sample size of at least n, in which n is found M = 0.04.

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.1*0.9}{n}}$

$0.04\sqrt{n} = 0.588$

$\sqrt{n} = \frac{0.588}{0.04}$

$\sqrt{n} = 14.7$

$(\sqrt{n})^{2} = (14.7)^{2}$

$n = 216$

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.

7 0

3 years ago

Walk fifty meters at 30o north or east from the old oak tree. (2) Turn 45o to your left (you should now be facing 75o north of e

saw5 [17]

Answer:

A straight line of approximately 75 meters, 1.4º north

Step-by-step explanation:

Hi, let's make it step by step to make it clearer

1) If we walk 50 meters in 30º angle Northeast, assuming the Old Oak tree is the point 0,0 and we're dealing with vectors in $R^{2}$ . To say 30º Northeast is 30º clockwise (or 60º counter clockwise).

2) Then there was a the turning point to the left. If I turn to the left, on my compass 45º , I'll face 75º northeast.

3) Finally, the last vector leads to the treasure from the Old Oak Tree, i.e. the resultant.

So, let's calculate the norm which is the length of the each vector.

1) Graphing them we can find the points, then the components and then calculate the norm, the length of each vector.

Since the Oak Tree is on (0,0). The turning point (50,86.61) and the Rock (R=(1.4,74,85) we can write the following vectors:

$\vec{u}=\left \langle 50,86.61 \right \rangle\\\vec{v}=\left \langle -48.6,-11.76\right \rangle\\\vec{w}=\left \langle 1.4,74.85 \right \rangle$

Now, let's calculate each vector length by calculating the norm.

$\left \| \vec{u} \right \|=\sqrt{50^{2}+86.6^2}=100\\\left \| \vec{v} \right \|=\sqrt{(-48.6)^2+(-11.76)^2}=50\\\left \| \vec{u} \right \|=\sqrt{(1.4)^2+(74.85)^2}=74.86$

The path is almost 75 meters. And since it is less than 15º degrees to the left of the North (or to the right) its direction is still north of the Old Oak Tree.