The ratio of blueberries and strawberries is 1:7. If there are 210 strawberries how many blueberries are there?

Paul started work at company B ten years ago at the salary of $30,000. At the same time Sharla started at Company C at the salar

Naily [24]

Answer:

Sharla has more, by $2000.

Step-by-step explanation:

Company B has 30000+2400x.

Company C has 36000+2000x.

Since we know x is equal to 10, because they have been working from the last ten years, we plug in 10 at where the x has been.

Company B: 30000+2400(10)

Company C: 36000+2000(10)

Let's first solve for B.

30000+2400(10)= 30000+24000=54000.

So, Paul earned $54000.

Next, Let's solve for C.

36000+2000(10)= 36000+20000=56000.

Sharla earned $56000.

It is obvious that sharla has more, so we will find the difference between the earnings of two companies.

56000-54000=2000.

So, Sharla has more earnings, with $2000 more than Paul.

5 0

2 years ago

A movie theater manager wants to determine whether popcorn sales have increased since the theater switched from using "butter-fl

labwork [276]

Answer:

The function that would correctly calculate the 90% range of likely sample means is given by:

B. 4,200±CONFIDENCE.T(0.10,140,12)

Step-by-step explanation:

In Microsoft Excel, the syntax

$CONFIDENCE.T(alpha,standard-dev,size)$ returns the confidence interval for the population mean, using the students T-distribution. $Alpha=1-90\%=10\%=0.10$

The standard deviation is given as $140 and the sample size is 12.

In constructing the confidence interval we use:

$\bar X\pm CONFIDENCE.T(alpha,standard-dev,size)$

Let us substitute the values to get:

$4,200\pm CONFIDENCE.T(0.10,140,12)$

We use the T distribution because $\sigma$ is unknown

8 0

2 years ago

A machine called Feynman, produces output, that is 5 times its input followed by an increase of 2. Another machine called Maxwel

olganol [36]

I honestly think that the answer is a

7 0

2 years ago

Read 2 more answers

What is the expression in radical form? (7x^3y^2) 2/7

Gelneren [198K]

$\sqrt[7]{(7x^3y^2)^{2} } = \sqrt[7]{(49x^6y^4)}$
In a fractional exponent the denominator tells what root it is (7th root in this case) and the numerator tells what the value in the parenthesis is raised to (2nd power in this case)

5 0

2 years ago

Read 2 more answers

Jeanne wants to start collecting coins and orders a coin collection starter kit. The kit comes with three coins chosen at random

lesya692 [45]

Conditional probability is a measure of the probability of an event given that another event has occurred. If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A|B), or sometimes $P_B(A)$ .

The conditional probability of event A happening, given that event B has happened, written as P(A|B) is given by
$P(A|B)= \frac{P(A \cap B)}{P(B)}$

In the question, we were told that there are three randomly selected coins which can be a nickel, a dime or a quarter.

The probability of selecting one coin is $\frac{1}{3}$

Part A:
To find the probability that all three coins are quarters if the first two envelopes Jeanne opens each contain a quarter, let the event that all three coins are quarters be A and the event that the first two envelopes Jeanne opens each contain a quarter be B.

P(A) means that the first envelope contains a quarter AND the second envelope contains a quarter AND the third envelope contains a quarter.

Thus $P(A)= \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}$

P(B) means that the first envelope contains a quarter AND the second envelope contains a quarter

Thus $P(B)= \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$

Therefore, $P(A|B)=\left( \frac{ \frac{1}{27} }{ \frac{1}{9} } \right)= \frac{1}{3}$

Part B:
To find the probability that all three coins are different if the first envelope Jeanne opens contains a dime, let the event that all three coins are different be C and the event that the first envelope Jeanne opens contains a dime be D.

$P(C)= \frac{3}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{6}{27} = \frac{2}{9}$

 $P(D)= \frac{1}{3}$ 

Therefore, $P(C|D)=\left( \frac{ \frac{2}{9} }{ \frac{1}{3} } \right)= \frac{2}{3}$

3 0

2 years ago