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3 years ago

What is the equation of the line through the origin and (3, 4)?

Mathematics

1 answer:

Lady_Fox [76]3 years ago

6 0

Answer:

A logarithm is the power to which a number must be raised in order to get some other number (see Section 3 of this Math Review for more about exponents). For example, the base ten logarithm of 100 is 2, because ten raised to the power of two is 100: log 100 = 2. because.

STep-by-step Ihave a felling its b

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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

Triangles ABC and DEF are similar

Sloan [31]

Answer:did U Get the answer or no

Step-by-step explanation:

Did u get the answers cuz I got the same question

7 0

3 years ago

Approximately 1,870,000 oz of gold went into the manufacturing of electronic equipment in a certain country in 1 year. This was

Alexxandr [17]

Answer: 17000000 ounces of gold were mined in that country that year.

Step-by-step explanation:

Let x represent the number of ounces of gold that were mined in that country that year.

Approximately 1,870,000 oz of gold went into the manufacturing of electronic equipment in a certain country in 1 year. This was 11% of all the gold mined in that country that year. This means that

11/100 × x = 1870000

0.11x = 1870000

x = 1870000/0.11

x = 17000000

7 0

3 years ago

A highway had a landslide, where 3,00 cubic yards of material fell on the road, requiring 200 dump truck loads to clear. On anot

Andru [333]

Answer:

2667 dump truck loads

Step-by-step explanation:

To solve this exercise, we must resort to a proportion rule.

Knowing what happened in the first place, how much was needed, a direct proportion will be made with the other accident, be as follows.

Landslide Dump Truck Loads

3 yd ^ 3 -------------------------------------------> 200

40 yd ^ 3 -------------------------------------------> X

X = (40 * 200) / 3

X = 2666.6, that is, for 40 cubic yards of sliding, approximately 2667 dump truck loads are needed.

8 0

3 years ago

rosa works north of her home. her husband works east they leave at the sane time. by the time rosa is 7 miles from home, the dis

Gennadij [26K]

This problem requires the Pythagorean therom.

The distance between them is 7.

Since the distance between them is one more mile than ray's distance from home ray is 6 miles from home.

So rosa is $\sqrt{7^2-6^2}$

So rosa is $\sqrt{13}$ miles away from home.

Rosa is about 3.6 miles from home.

7 0

3 years ago