The temperature is 45°F The temperature will decrease by 2°F each hour. Let

Sarah answered 24 questions on her test correctly and earned an 80%. How many questions were on the test?

MaRussiya [10]

Answer:

43 questions in total

Step-by-step explanation:

the original equation with the unknown (total number of questions on the test) will look like this

$\frac{total \: number \: of \: questions \: on \: exam}{24 \: answered \: questions} = 80\%$

we know this because to work out a percentage in the 1st place you need the total ÷ number (in this case it's ÷ how many questions out of the total she did answer)

now you rearrange.

you need to move the 24 to get the total on it's own here's how you do it

$\frac{tot \: num \: of \: qs}{24} \times 24 = 80\% \times 24$

what you do to one side of equation you have to do to the other. the 24 and 24 on the left will cancel because 24÷24=1 and we don't write

got mum of Q's x 1= 80 x 24

now you convert % to decimal

$80\% \times 100 = 0.80$

replace

$tot \: num \: qs \: = 24 \times 0.80 = 19$

now you take the original and + the 19

$24 + 19 = 43$

therefore there are a total of 43 questions on the exam

5 0

4 years ago

0.5 is 5 % of what number ?

Mrrafil [7]

It would be .10 bc they are both decimals

5 0

4 years ago

Read 2 more answers

A box of Great Value crackers is $1.97 and contains 27 servings. What is the price per EACH serving?

nlexa [21]

.07 dollars or 7 cents

4 0

3 years ago

Use the substitution of x=e^{t} to transform the given Cauchy-Euler differential equation to a differential equation with consta

kherson [118]

By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}\implies\dfrac{\mathrm dy}{\mathrm dt}=x\dfrac{\mathrm dy}{\mathrm dx}$

which follows from $x=e^t\implies t=\ln x\implies\dfrac{\mathrm dt}{\mathrm dx}=\dfrac1x$ .

$\dfrac{\mathrm dy}{\mathrm dt}$ is then a function of $x$ ; denote this function by $f(x)$ . Then by the product rule,

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dt}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dt}+\dfrac1x\dfrac{\mathrm df}{\mathrm dx}$