Supplementary angle of 80

K is the midpoint of JL JK=3x-1 and KL=2x+8 find JL

Masteriza [31]

Answer:

JL = 52

Step-by-step explanation:

It is given in the question that K is the midpoint of JL.

So, JK = KL ......... (1)

Now, given that, JK =3x - 1 and KL = 2x + 8

Therefore, from equation (1), we can write

3x - 1 = 2x + 8

⇒ x = 9

So, JK = 3x - 1 = 3(9) -1 = 26 and KL = 2x + 8 = 2(9) + 8 = 26

Hence, JL = JK + KL = 26 + 26 = 52. (Answer)

3 0

3 years ago

Y varies directly to some quantities and varies inversely to some other quantities

yulyashka [42]

Answer:

I don't understand your question

Step-by-step explanation:

5 0

3 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

(c) 90.53%

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .

$p_2=0.94$

The total number of shipment, i.e sample size, $n_2= 300$

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, $\mu_2$ , and variance, $\sigma_2^2$ .

Mean: $\mu_2=n_2p_2=300\times0.94=282$

Variance: $\sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92$

$\Rightarrow \sigma_2=\sqrt(16.92}=4.11$ .

In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85$ .

So, the probability that a given shipment is acceptable is

$P(z\geq0.85)=\int_{0.85}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}=0.1977$

Hence, the probability that a given shipment is acceptable is 0.20.

(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

For this value, the value of z-score is -3.09 (from the z-score table)

Let, $\alpha$ be the required probability of acceptance of one shipment.

So,

$-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}$

On solving

$\alpha= 0.977896$

Again, the probability of acceptance of one shipment, $\alpha$ , depends on the probability of meeting the thickness specification of one bearing.

For this case,

The area right to the z-score=0.97790

and the area left to the z-score is 1-0.97790=0.0221.

The value of z-score is -2.01 (from the z-score table)

Let $p$ be the probability that one bearing meets the specification. So

$-2.01=\frac{439.5-500 p}{\sqrt{500 p(1-p)}}$

On solving

$p=0.9053$

Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.

8 0

4 years ago

What is the answer? please help

amid [387]

Answer:

solution,

A(2,7)---------> (X1,y1)

B(6,3)--------->(x2,y2)

now,

 $ab = ( \frac{x1 + x1}{2} \: , \frac{y1 + y2}{2} ) \\ = (\frac{2 + 6}{2} \: , \frac{7 + 3}{2} ) \\ = ( \frac{8}{2} \:, \frac{10}{2} ) \\ = (4 ,\: 5)$ 

hope this helps...

Good luck on your assignment...

7 0

3 years ago

Read 2 more answers

Which case allows for more than one triangle with the given measures to be constructed?

Anni [7]

A.three sides measuring 9 meters, 5 meters, and 5 meters

--> only one isosceles triangle

B.three sides measuring 6 inches, 3 inches and 9 inches

--> none triangle may be formed because 9 = 3 + 6 and there is a rule (the triangle inequality theorem) that states the length of any side must be less than the sum of the lengths of other two sides.

C.three angles measuring 25°, 25°, and 130°

--> 25 + 25 + 130 = 180 => you can construct many triangles with these angles

D.three angles measuring 55°, 25°, and 125°

--> 55 + 25 + 125 = 205 > 180 => you cannot construct any triangle with theses measures.

7 0

4 years ago