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Oksi-84 [34.3K]

3 years ago

6

Name three pairs of numbers that have 5 as their gfc (Greatest Common factor). Use each number once in your answer. I really nee

d help asap. Thanks! (:

2 answers:

katen-ka-za [31]3 years ago

4 0

20,25 and 35,50 and 10,15. Each pair would have a GCF of 5.

Fed [463]3 years ago

3 0

Any three numbers that are multiples of 5 can be your answer. i would use 5 10 and 15

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Find the nth term:

san4es73 [151]

Answer:

n+3

Step-by-step explanation:

the relationship between those numbers are consistently adding 3 thus n +3

7 0

3 years ago

Read 2 more answers

3a+3b,for a=2 and b=4

il63 [147K]

3a + 3b, for a=2 and b=4 ... The first thing you'd do would be to plug the numbers in -

3 x 2 + 3 x 4

Then multiply them -

6 + 12

Then add -

18.

So your answer should be 18.

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

3 years ago

Bruce pitches for his little brother's baseball team. He has observed that the number of pitches a batter hits varies and is giv

liberstina [14]

Hello!

If Bruce observes that the number of pitches a batter hits varies and is given by the function f(x)=x-11, and the batters get {4, 12, 14, 27, 42}, then Bruce threw {15, 23, 25, 38, 53} pitches. We get this solution set by adding 11 to each element in the set {4, 12, 14, 27, 42}.
Have a nice day

6 0

3 years ago

A cylinder has a base radius of 7m and a height of 14m. What is its volume in cubic m, to the nearest tenths place?

Westkost [7]

Answer:Answer:

2155.1 m^3

Step-by-step explanation:

area= pi* r ^2 * h =pi * 7^2 *14 = 2155.1 m^3

6 0

2 years ago