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1 year ago

A=V1-V0/t solve for V1

2 answers:

8 0

For this case we have the following equation:
$A = \frac{v1-v0}{t}$
From here, we must clear the value of v1.
For this, we follow the following steps:
1) Pass variable t to multiply:
$A*t = v1 - v0$
2) Add v0 on both sides of the equation:
$A*t+v0=v1-v0+v0$
3) Rewrite the expression:
$A*t+v0=v1$
Answer:
The resolved expression for v1 is:
$v1 = A*t+v0$

MA_775_DIABLO [31]1 year ago

6 0

A = V1 - V0/t.....multiply both sides by t
At = V1 - V0....add V0 to both sides
At + V0 = V1

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Find the volume of a pyramid with a square base, where the perimeter of the base is 5.1 in 5.1 in and the height of the pyramid

lana66690 [7]

Step-by-step explanation:

PARA ENCONTRAR CUÁNTO MIDE CAD LADO SE DIVIDE EL PERÍMETRO

ENTRE 4 LADOS

l = 1.28 in

LOEGO ENCUENTRAS EL RADIO DE LA PIRÁMIDE

<em>r</em> = 2.55 in

USANDO LA FÓRMULA

V = 4(1.28 in)(2.55 in)(2.7 in)/6 = $5.9 in^{3}$

6 0

2 years ago

J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.

melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean for mail-order sales = $74.50

$\bar X_2$ = sample mean for internet sales = $84.40

$s_1$ = sample standard deviation for mail-order purchases = $17.25

$s_2$ = sample standard deviation for internet purchases = $21.25

$n_1$ = sample of sales receipts for mail-order purchases = 16

$n_2$ = sample of sales receipts for internet purchases = 9

Also, $s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} }$ = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by ( $\mu_1-\mu_2$ ).

Now, 99% confidence interval for ( $\mu_1-\mu_2$ ) is given by;

= $(\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}$

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

= $(74.50-84.40) \pm (2.807 \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})$

= [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

5 0

1 year ago

What is the distance between (3, 5.25) and (3, –8.75)? 6 units 8.25 units 11.75 units 14 units

evablogger [386]

By using distance formula :

$\text{Distance formula,} \bold{ \boxed{ Distance = \sqrt{( x_{2} -x_{1})^{2}+(y_{2} - y_{1})^{2}) }}}$

Given points = ( 3 , 5.25 ) and ( 3 , - 8.75 )

$\bold{Taking \: \: \: x_{1}=3 \: \: , \: \: x_{2}= 3 \: \: , \: \: y_{1}= 5.25 \: \: , \: \: y_{2}= -8.75}$

On applying formula, we get

$Distance = \sqrt{ ( x_{2}-x_{1})^{2}+(y_{2}-y_{1})^2} \\ \\ \\ Distance = \sqrt{ ( 3 - 3 )^{2} + ( - 8.75 - 5.25 )^{2}} \\ \\ \\ Distance = \sqrt{ ( 0 )^{2} + ( - 14)^{2}} \\ \\ \\ Distance = \sqrt{ ( - 14 )^{2}} \\ \\ \\ Distance = \sqrt{ 14^{2}} \:\:\:\:\:\:\:\:\:\:\: \: \: \: \: \: \: \: \: | \bold{ ( - 14 )^{2} = 14^{2}} \\ \\ \\ Distance = {14}^{2 \times \frac{1}{2} } \\ \\ \\ Distance = {14}^{1} \\ \\ \\ Distance = 14 \: units$

Hence, Option D is correct.

4 0

2 years ago

Read 2 more answers

A rectangular tank measuring 35 cm by 28 cm by 16 cm is 2/5 filled with

Alina [70]

Answer:

Amount of water left in tank = 440 cm³

Step-by-step explanation:

Given:

Sides of tank respectively = 35 cm , 28 cm , 16 cm

Water level in tank = 2/5

Side of a cubic tank = 18 cm

Find:

Amount of water left in tank

Computation:

Volume level in rectangular tank = [2/5][35 x 28 x 16]

Volume level in rectangular tank = [2/5][15,680]

Volume level in rectangular tank = [31,360/5]

Volume level in rectangular tank = 6,272 cm³

Volume of cube = side³

Volume of cube = 18³

Volume of cube = 5,832 cm³

Amount of water left in tank = Volume level in rectangular tank - Volume of cube

Amount of water left in tank = 6,272 - 5,832

Amount of water left in tank = 440 cm³

7 0

1 year ago

Flip two coins 100 times, and record the results of each coin toss in a table like the one below:

monitta

Answer:

1)The theoretical probability that a coin toss results in two heads showing is 25%.

2)The experimental probability that a coin toss results in two heads showing is 44%.

3) The theoretical probability that a coin toss results in two tails showing is 25%.

4) The experimental probability that a coin toss results in two tails showing is 34%.

5) The theoretical probability that a coin toss results in one head and one tail showing is 50%.

6) The experimental probability that a coin toss results in a head and a tail is 22%.

7) The experimental probabilities are slightly different from the theoretical probabilities because the number of experiments is relatively small. As the number of experiments increase, the experimental probabilities will get closer to the theoretical probabilities.

Step-by-step explanation:

Probability:

What you want to happen is the desired outcome.

Everything that can happen iis the total outcomes.

The probability is the division of the number of possible outcomes by the number of total outcomes.

Theoretical Probability:

The results you expect to happen.

Experimental Probability:

The probability determined from the result of an experiment.

1. What is the theoretical probability that a coin toss results in two heads showing?

In each toss, the theoretical probability that a coin toss results in a head showing is 50%.

So for two coins, the probability is:

$P = (0.5)^{2} = 0.25$

The theoretical probability that a coin toss results in two heads showing is 25%.

2. What is the experimental probability that a coin toss results in two heads showing?

There were 100 flips, and it resulted in two heads 44 times, so:

$P = \frac{44}{100} = 0.44$

The experimental probability that a coin toss results in two heads showing is 44%.

3. What is the theoretical probability that a coin toss results in two tails showing?

In each toss, the theoretical probability that a coin toss results in a tail showing is 50%.

So for two tails, the probability is:

$P = (0.5)^{2} = 0.25$

The theoretical probability that a coin toss results in two tails showing is 25%.

4. What is the experimental probability that a coin toss results in two tails showing?

There were 100 flips, and it resulted in two tails 34 times, so:

$P = \frac{34}{100} = 0.34$

The experimental probability that a coin toss results in two tails showing is 34%.

5. What is the theoretical probability that a coin toss results in one head and one tail showing?

In each toss, the theoretical probability that a coin toss results in a tail showing is 50% and in a head showing is 50%.

They can be permutated, as the tail can appear before the head, or the head before the tail. So:

$P = p_{2,1}*(0.5)*(0.5) = \frac{2!}{1!}*0.25 = 0.50$

The theoretical probability that a coin toss results in one head and one tail showing is 50%.

6. What is the experimental probability that a coin toss results in one head and one tail showing?

There were 100 flips, and it resulted in a head and a tail showing 22 times, so:

$P = \frac{22}{100} = 0.22$

The experimental probability that a coin toss results in a head and a tail is 22%.

6 0

1 year ago