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3 years ago

Suppose , varies jointly with g and v, and j = 2 when g = 4 and v= 3.

Mathematics

1 answer:

kap26 [50]3 years ago

7 0

Answer:

j = 44/3

Step-by-step explanation:

j varies jointly as g and v. This can be represented mathematically as:

$j \alpha gv\\j = kgv$ .............(1)

Where k is a constant of proportionality

j = 2 when g = 4 and v = 3

Substitute these values into equation (1)

2 = k * 4 * 3

2 = 12 k

k = 1/6

when g = 8 and v = 11:

j = (1/6) * 8 * 11

j = 44/3

You might be interested in

Past experience has indicated that the breaking strength of yarn used in manufacturing drapery material is normally distributed

vlabodo [156]

Answer:

95% two-sided confidence interval on the true mean breaking strength is (94.8cm, 99.2cm)

Step-by-step explanation:

Our sample size is 11.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

$df = 11-1 = 10$ .

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 10 and 0.025 in the two-sided t-distribution table, we have $T = 1.812$

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

$s = \frac{4}{\sqrt{11}} = 1.2060$

Now, we multiply T and s

$M = Ts = 1.812*1.2060 = 2.19/tex]For the lower end of the interval, we subtract the sample mean by M. So the lower end of the interval here is[tex]L = 97 - 2.19 = 94.81 = 94.8$ cm

For the upper end of the interval, we add the sample mean and M. So the upper end of the interval here is

$L = 97 + 2.19 = 99.19 = 99.2$ cm

95% two-sided confidence interval on the true mean breaking strength is (94.8cm, 99.2cm).

7 0

2 years ago

Which methods could you use to calculate the x-coordinate of the midpoint of

Iteru [2.4K]

Which methods could you use to calculate the x-coordinate of the midpoint of

a horizontal line segment with endpoints at (0,0) and (20, 0)?

Check all that apply.

C. Count by hand

3 0

3 years ago

PLEASE HELP!!!:)). The area of the trampezoid is 100dm and the bases are 2.1m and 1.9m. Calculate the height of the trampezoid

mixas84 [53]

Answer:

Well, the only thing you should do is to use the formula.

if the bases are: a, b

and the height is=h

Then, this is your formula, S=½(a+b)×h

Step-by-step explanation:

Aight,

100dm=10m

S=10m

now, the formula

10=½(2.1+1.9)×h===> 20=4h==> h=5m

5 0

2 years ago

Akira and her puppy are both dressing up as superheroes for Halloween this year. Akira needs 1 yard of cloth for her cape. She a

Ahat [919]

The feet of cloth that she should buy for the capes is 4 feet.

<h3>How many feet should she buy?</h3>

The mathematical operations that would be used to determine the answer are addition and multiplication.

Addition is the process of summing two or more numbers together. Multiplication is the process of determining the product of two or more numbers.

The first step is to convert yard to feet

According to the unit of conversion : 1 yard = 3 feet

1 x 3 = 3 feet

The second step is to add the feet needed for Akira's cape with the foot needed for the puppy's cape

Total feet needed = 3 + 1 = 4 feet

To learn more about addition, please check: brainly.com/question/349488

#SPJ1

8 0

1 year ago

A real estate agent has 19 properties that she shows. She feels that there is a 30% chance of selling any one property during a

netineya [11]

Answer:

$P(X \geq 5)=1-P(X$

We can find the individual probabilities:

$P(X=0)=(19C0)(0.3)^0 (1-0.3)^{19-0}=0.00114$

$P(X=1)=(19C1)(0.3)^1 (1-0.3)^{19-1}=0.0092$

$P(X=2)=(19C2)(0.3)^2 (1-0.3)^{19-2}=0.0358$

$P(X=3)=(19C3)(0.3)^3 (1-0.3)^{19-3}=0.0869$

$P(X=4)=(19C4)(0.3)^4 (1-0.3)^{19-4}=0.1491$

And replacing we got:

$P(X \geq 5) = 1-[0.00114+0.009282+0.0358+0.0869+0.149]= 0.7178$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=19, p=0.3)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

And we want to find this probability:

$P(X \geq 5)$

And we can use the complement rule:

$P(X \geq 5)=1-P(X$

We can find the individual probabilities:

$P(X=0)=(19C0)(0.3)^0 (1-0.3)^{19-0}=0.00114$

$P(X=1)=(19C1)(0.3)^1 (1-0.3)^{19-1}=0.0092$

$P(X=2)=(19C2)(0.3)^2 (1-0.3)^{19-2}=0.0358$

$P(X=3)=(19C3)(0.3)^3 (1-0.3)^{19-3}=0.0869$

$P(X=4)=(19C4)(0.3)^4 (1-0.3)^{19-4}=0.1491$

And replacing we got:

$P(X \geq 5) = 1-[0.00114+0.009282+0.0358+0.0869+0.149]= 0.7178$

4 0

3 years ago