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2 years ago

Rearrange the formula d = m/v for v

2 answers:

6 1

D = m/v

v*d = v*(m/v) ... multiply both sides by v

v*d = m

(v*d)/d = m/d ... divide both sides by d

v = m/d

--------------------------------

After we solve for v (aka isolate v), we end up with this final answer

v = m/d

Guest

1 year ago

This is wrong.

postnew [5]2 years ago

7 0

Answer:

v=m/d

Step-by-step explanation:

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Rewrite the following expression using the properties of logarithms. log2z+2log2x+4log9y+12log9x−2log2y

kramer

Use the rules of logarithm:
1. log(x)+log(y)=log(xy)
log(x)-log(y)=log(x/y)
2. k*log(x) = log(x^k)
log(x)/k = log(x^(1/k))

<span>log(2z)+2log(2x)+4log(9y)+12log(9x)−2log(2y)
=</span>log(2z)+log(4x^2)+log(9^4y^4)+log(9^12x^12)−log(4y^2)
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5 0

2 years ago

Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by $2P(T$ 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

where $t_{0.025}$ is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

8 0

2 years ago

The base of a triangle measures 8 inches and the area is 136 square inches what is the height of the triangle

Mariana [72]

Answer:

34 inches

Step-by-step explanation:

Area is base times height time half

Then plug in the values: 136= 0.5 x 8x

Then solve for x

7 0

2 years ago

According to a study in a medical journal, 202 of a sample of 5,990 middle-aged men had developed diabetes. It also found that m

tekilochka [14]

Answer:

0.0588 = 5.88% probability that a middle-aged man with diabetes is very active

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Has diabetes.

Event B: Is very active.

Probability of having diabetes:

To find this probability, we take in consideration that:

It also found that men who were very active (burning about 3,500 calories daily) were a fourth as likely to develop diabetes compared with men who were sedentary. Assume that one-fifth of all middle-aged men are very active, and the rest are classified as sedentary.

So the probability of developing diabetes is:

x of 4/5 = x of 0.8(not active)

x/4 = 0.25x of 1/5 = 0.2(very active). So

$P(A) = 0.8x + 0.25*0.2x = 0.85x$

Probability of developing diabetes while being very active:

0.25x of 0.2. So

$P(A \cap B) = 0.25x*0.2 = 0.05x$

What is the probability that a middle-aged man with diabetes is very active?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.05x}{0.85x} = \frac{0.05}{0.85} = 0.0588$

0.0588 = 5.88% probability that a middle-aged man with diabetes is very active

4 0

2 years ago

Find the following measure for this figure. Lateral area = 24 square units 48 square units 96 square units

Savatey [412]

We are asked to solve for the lateral surface area of the cylinder such that the formula that we can used is Lateral Surface Area = 2pi*r*h where "r" represents the radius of the cylinder and "h" represents the height of the cylinder. In this problem, we are given with the following values:
r = 4 inches
h = inches
Solving for the lateral surface area, we have it:
L.S.A = 2 *pi * r* h
L.S.A = 2*pi*4*6
L.S.A = 48pi units squared

The answer is 48pi unit2.

8 0

2 years ago

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