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2 years ago

Triangle not drawn to scale

Mathematics

1 answer:

Liono4ka [1.6K]2 years ago

5 0

Answer:

D. scalene triangle

Step-by-step explanation:

This triangle has sides that all have different lengths. Therefore, this triangle is a scalene triangle.

Tips:

Isosceles triangle - two sides have equal lengths

Equilateral triangle - all three sides have equal lengths

I hope this helped! :)

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Y = 1/2x + 6 Y = -2x + 3 Y= 1/2x + 3 Y= -1/2x + 3

TEA [102]

Answer:

y=-1/2x+3

Step-by-step explanation:

it has a negative slope and it's not that steep which is why the slope isnt 2. also the first one is wrong because the y-intercept is 3

7 0

3 years ago

Read 2 more answers

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

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

3 years ago

How do you math please help me

Stella [2.4K]

Math is math math gives more math math is math math is more math more math is tons of math tons of math is big brain math knowledge big brain math knowledge means your math

3 0

3 years ago

Can somebody help me with this please

dmitriy555 [2]

Uhhh, kind of an estimated guess but I got $7426.30

4 0

3 years ago

Is 9.521521521 rational or irrational

enyata [817]

If the numbers after the decimal terminate, yes, it's rational.

9.521521521 = 9,521,521,521 / 1,000,000,000

If they don't terminate, but the pattern continues (which I suspect is the case here), yes, it's still rational.

If x = 9.521521521…, then

1000x = 9521.521521521…

Subtract x from this to eliminate the fractional part:

1000x - x = 9521.521521521… - 9.521521521…

999x = 9512

x = 9512/999

If they don't terminate, but the pattern does not continue, meaning the next few digits could be something random like

9.52152152119484929271283583457…

then the number would be irrational.

7 0

3 years ago