we are given
Firstly , we can distribute
now, we can combine like terms
now, we can solve for x
and we get
...........Answer
9. Bella hiked a 13-kilometer trail at Great Peak. How many meters did Belle
1 answer:
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Answer:
<em>AD is 6.5 units</em>
Step-by-step explanation:
Find the diagram attached.
From the diagram, you can see that the dotted triangle is a right angled triangle with side AD as the hypotenuse.
To get AD, we will use the pythagoras theorem as shown;
Hyp² = Opp² + Adj²
AD² = 6²+2.5²
AD² = 36 + 6.25
AD² = 42.25
AD = √42.25
<em>AD = 6.5</em>
<em>Hence the measure of length segment AD is 6.5 units</em>
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 be the probability that a bearing meets the specification.
So,
Sample size, , is large.
Let X represent the number of acceptable bearing.
Convert this to a normal distribution,
Mean:
Variance:
(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.
So,
Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.
.
So, the probability that a given shipment is acceptable is
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 .
The total number of shipment, i.e sample size,
Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, , and variance,
.
Mean:
Variance:
.
In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.
.
So, the probability that a given shipment is acceptable is
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, be the required probability of acceptance of one shipment.
So,
On solving
Again, the probability of acceptance of one shipment, , 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 be the probability that one bearing meets the specification. So
On solving
Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.