All categories

2 years ago

(02.06)If 1 pint is equal to 2 cups, then 5 pints would equal ___ cups.

Mathematics

1 answer:

EleoNora [17]2 years ago

3 0

Answer:

If 1 pint is equal to 2 cups, then 5 pints would equal _<u>5</u>_ cups.

Step-by-step explanation:

Let's use a ratio to solve:

pints : cups

1 : 2

2 : 4

3 : 6

4 : 8

5 : 10

There would be 10 cups in 5 pints.

You might be interested in

PLAESE HELP ASAP!!!!

vovikov84 [41]

11 5/8 + 9 1/2= 21 1/8
3/4 - 2/5= 7/20
4 5/6 - 2 1/2= 2 1/3
I think these are right, I apologize if they are wrong.

5 0

2 years ago

Read 2 more answers

Need help pls<br>thank you in advance

ira [324]

Answer:

<u />

$\textsf{Mean}\:\overline{X}=\sf \dfrac{\textsf{sum of all the data values}}{\textsf{total number of data values}}$

$\implies \sf Mean\:(Nilo)=\dfrac{5+6+14+15}{4}=\dfrac{40}{4}=10$

$\implies \sf Mean\:(Lisa)=\dfrac{8+9+11+12}{4}=\dfrac{40}{4}=10$

<h3><u>Standard Deviation</u></h3>

$\displaystyle \textsf{Standard Deviation }s=\sqrt{\dfrac{\sum X^2-\dfrac{(\sum X)^2}{n}}{n-1}}$

$\begin{aligned}\displaystyle \textsf{Standard Deviation (Nilo)} & =\sqrt{\dfrac{(5^2+6^2+14^2+15^2)-\dfrac{(5+6+14+15)^2}{4}}{4-1}}\\\\& = \sqrt{\dfrac{482-\dfrac{40^2}{4}}{3}}\\\\& = \sqrt{\dfrac{82}{3}}\\\\& = 5.23\end{aligned}$

$\begin{aligned}\displaystyle \textsf{Standard Deviation (Lisa)} & =\sqrt{\dfrac{(8^2+9^2+11^2+12^2)-\dfrac{(8+9+11+12)^2}{4}}{4-1}}\\\\& = \sqrt{\dfrac{410-\dfrac{40^2}{4}}{3}}\\\\& = \sqrt{\dfrac{10}{3}}\\\\& = 1.83\end{aligned}$

<h3><u>Summary</u></h3>

Nilo has a mean score of 10 and a standard deviation of 5.23.

Lisa has a mean score of 10 and a standard deviation of 1.83.

The <u>mean</u> scores are the <u>same</u>.

Nilo's standard deviation is higher than Lisa's. Therefore, Nilo's test scores are more <u>spread out</u> that Lisa's, which means Lisa's test scores are more <u>consistent</u>.

5 0

1 year ago

F(x+1)+f(x+2)=2x+3; f(x)=?

Rama09 [41]

Answer:

$F=\frac{-xf+f+2x}{x+1};\quad \:x\ne \:-1$

Step-by-step explanation:

$1. \mathrm{Subtract\:}f\left(x+2\right)\mathrm{\:from\:both\:sides}\\
F\left(x+1\right)+f\left(x+2\right)-f\left(x+2\right)=2x+3f-f\left(x+2\right)

$

$2. \mathrm{Simplify}\\
F\left(x+1\right)=-xf+f+2x\\
$

$3. \mathrm{Divide\:both\:sides\:by\:}x+1;\quad \:x\ne \:-1\\
\frac{F\left(x+1\right)}{x+1}=-\frac{xf}{x+1}+\frac{f}{x+1}+\frac{2x}{x+1};\quad \:x\ne \:-1$

$4. \mathrm{Simplify}\\
F=\frac{-xf+f+2x}{x+1};\quad \:x\ne \:-1$

Final Answer: $F=\frac{-xf+f+2x}{x+1};\quad \:x\ne \:-1$

4 0

2 years ago

Add an intersection the red light times normally distributed by the mean of three minutes and a standard deviation of .25 minute

Soloha48 [4]

95% of red lights last between 2.5 and 3.5 minutes.

<u>Step-by-step explanation:</u>

In this case,

The mean M is 3 and
The standard deviation SD is given as 0.25

Assume the bell shaped graph of normal distribution,

The center of the graph is mean which is 3 minutes.

We move one space to the right side of mean ⇒ M + SD

⇒ 3+0.25 = 3.25 minutes.

Again we move one more space to the right of mean ⇒ M + 2SD

⇒ 3 + (0.25×2) = 3.5 minutes.

Similarly,

Move one space to the left side of mean ⇒ M - SD

⇒ 3-0.25 = 2.75 minutes.

Again we move one more space to the left of mean ⇒ M - 2SD

⇒ 3 - (0.25×2) =2.5 minutes.

The questions asks to approximately what percent of red lights last between 2.5 and 3.5 minutes.

Notice 2.5 and 3.5 fall within 2 standard deviations, and that 95% of the data is within 2 standard deviations. (Refer to bell-shaped graph)

Therefore, the percent of red lights that last between 2.5 and 3.5 minutes is 95%

8 0

3 years ago

I need help with this question

densk [106]

Answer:

So we have an original figure with a value of 50 and then a scaled version or dilation

Hence we have a scale factor of

$\frac{1}{5}$

3 0

2 years ago