What is 2 divided by the square root of zero

a buret is a tool designed to transfer precise amounts of liquid .A buret initially contains 70.00 millimeters of a solution and

katrin [286]

Let:

Vbu= Volume of the buret

Vbk= Volume of the beaker

A buret initially contains 70.00 millimeters of a solution and a beaker initially contains 20.00 ml of the solution the buret drips solution into the Beaker. each drip contains 0.05 mL of solution, therefore:

x = Number of drips

a = volume of each drip

$\begin{gathered} Vbu=70-ax \\ Vbk=20+ax \\ \text{where:} \\ a=0.05 \\ Vbu=70-0.05x \\ Vbk=20+0.05x \end{gathered}$

after how many drips will the volume of the solution in the buret and beaker be equal ? Vbu = Vbk:

$\begin{gathered} Vbu=Vbk \\ 70-0.05x=20+0.05x \\ \text{Solve for x:} \\ 0.1x=70-20 \\ 0.1x=50 \\ x=\frac{50}{0.1} \\ x=500 \end{gathered}$

5 0

1 year ago

Completely factor the given expression over the integers.<br><br> (Correct gets brainliest)

elena-14-01-66 [18.8K]

Answer:

50

Step-by-step explanation:

would it be 50 because you the q to the power of 2 is 2q. Then you would subtract to the other side of the equation so it would be -100=-2q. After you would divided the -2 in -2q on both sides and -100/-2 is 50. So your answer is q=50

7 0

3 years ago

What are the values of x and y in PQRS? What are PR and SQ?

mamaluj [8]

Step-by-step explanation:

sol;

x+1=y...(1)

3y-7=2x....(2)

or, 3(x+1)-72x [from (1)]

or, 3x+3-7=2x

or, 3x-2x=7-3

x=4

now,

putting the value of x in (1)

y=x+1

=4+1

=5

PR and SQ are the diagonal of PQRS.

4 0

2 years ago

Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean

Mamont248 [21]

Answer:

a) The percentage of snails that take more than 60 hours to finish is 4.75%.

b) The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c) The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d) 0% probability that a randomly-chosen snail will take more than 76 hours to finish

e) To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f) The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 50, \sigma = 6$

a. The percentage of snails that take more than 60 hours to finish is

This is 1 subtracted by the pvalue of Z when X = 60.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

1 - 0.9525 = 0.0475

The percentage of snails that take more than 60 hours to finish is 4.75%.

b. The relative frequency of snails that take less than 60 hours to finish is

This is the pvalue of Z when X = 60.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c. The proportion of snails that take between 60 and 67 hours to finish is

This is the pvalue of Z when X = 67 subtracted by the pvalue of Z when X = 60.

X = 67

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{67 - 50}{6}$

$Z = 2.83$

$Z = 2.83$ has a pvalue 0.9977

X = 60

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

0.9977 - 0.9525 = 0.0452

The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d. The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)

This is 1 subtracted by the pvalue of Z when X = 76.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{76 - 50}{6}$

$Z = 4.33$

$Z = 4.33$ has a pvalue of 1

1 - 1 = 0

0% probability that a randomly-chosen snail will take more than 76 hours to finish

e. To be among the 10% fastest snails, a snail must finish in at most hours.

At most the 10th percentile, which is the value of X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 50}{6}$

$X - 50 = -1.28*6$

$X = 42.32$

To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f. The most typical 80% of snails take between and hours to finish.

From the 50 - 80/2 = 10th percentile to the 50 + 80/2 = 90th percentile.

10th percentile

value of X when Z has a pvalue of 0.1. So X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 50}{6}$

$X - 50 = -1.28*6$

$X = 42.32$

90th percentile.

value of X when Z has a pvalue of 0.9. So X when Z = 1.28

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 50}{6}$

$X - 50 = 1.28*6$

$X = 57.68$

The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

5 0

3 years ago

The interest on $2,000 for 2 years is $320. What is the simple interest rate?

umka2103 [35]

R = 8%/year

Equation:
r = (1/t)(A/P - 1)
Calculation:
Solving our equation:
r = (1/2)((2320/2000) - 1) = 0.08
r = 0.08

Converting r decimal to R a percentage
R = 0.08 * 100 = 8%/year

The interest rate required to get a total amount, principal plus interest, of $2,320.00 from simple interest on a principal of $2,000.00 over 2 years is 8% per year.

7 0

2 years ago