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Oksi-84 [34.3K]

3 years ago

13

in the 6th grade, 48 students brought a ham and cheese sandwich for lunch. if this number represents 32% of the total of student

s in the 6th grade, how many total students are in the 6th grade?

1 answer:

Bas_tet [7]3 years ago

5 0

Answer: 150

Step-by-step explanation:

you can easily calculate 48 is 32 percent of what number by using any regular calculator, simply enter 48 × 100 ÷ 32 and you will get your answer which is 150

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11. Mr Lee bought a second hand car for

Marrrta [24]

Answer :

Depends on the rate compound interest is accrued. See answers.

Step-by-step explanation:

We need the compound interest formula which is:

$A = P (1+\frac{r}{n} )^{nt}$

A = final amount

P = initial principal balance

r = interest rate

n = number of times interest applied per time period

t = number of time periods elapsed

It doesn't say how frequently the interest is compounded, so we will do monthly, quarterly, and yearly.

MONTHLY

So we know the car costs $25,480 and he made a down payment of $10,000. By subtracting the down payment from the purchase price we find the loan amount.

$25,480 - $10,000 = $15,480

P = 15,480

r = 4.75% = .0475

n = 12 (12 months in a year)

t = 2

Plug everything into the compound interest formula.

$A = P (1+\frac{r}{n} )^{nt}$

$A = 15480(1+\frac{.0475}{12})^{12*2}$

$A = 15480(1+\frac{.0475}{12})^{24}$

$A = 15480(1+.00396)^{24}$

$A = 15480(1.00396)^{24}$

$A = 15480*1.0992$

$A = $17016.14$

Mr. Lee paid $17,016.14 when the bill came due at 2 years with interest compounded monthly.

QUARTERLY

P = 15,480

r = 4.75% = .0475

n = 4 (4 quarters in a year)

t = 2

$A = P (1+\frac{r}{n} )^{nt}$

$A = 15480(1+\frac{.0475}{4})^{4*2}$

$A = 15480(1+\frac{.0475}{4})^{8}$

$A = 15480(1+.0119})^{8}$

$A = 15480(1.0119)^{8}$

$A = 15480*1.099$

$A = 15480(1.099)$

$A = 17013.20$

Mr. Lee paid $17,013.20 when the bill came due at 2 years with interest compounded quarterly.

YEARLY

P = 15,480

r = 4.75% = .0475

n = 1

t = 2

$A = P (1+\frac{r}{n} )^{nt}$

$A = 15480(1+\frac{.0475}{1})^{1*2}$

$A = 15480(1+\frac{.0475}{12})^{2}$

$A = 15480(1+.0475})^{2}$

$A = 15480(1.0475)^{2}$

$A = 15480*1.0973$

$A = 16985.53$

Mr. Lee paid $16,985.53 when the bill came due at 2 years with interest compounded yearly.

4 0

4 years ago

Joey had $328. He found $5 in his pocket while walking to school. What was his percent increase?

Alex73 [517]

Answer:

his percentage was increased by 5

8 0

3 years ago

Read 2 more answers

What is the slope of the line that passes through (-3,4) and (-3,-4)

suter [353]

Answer:

undefined

Step-by-step explanation:

We can use the slope formula to find the slope

m = ( y2-y1)/(x2-x1)

= ( -4-4)/( -3 - -3)

= ( -4-4)/( -3+3)

= -8/0

When we divide by zero, the answer is undefined so the slope is undefined

8 0

3 years ago

A particle moves in a straight line so that its velocity at time

riadik2000 [5.3K]

Answer:

s(2) = 7.75

Step-by-step explanation:

given the velocity v(t) = t^3

we can find the position s(t) by simply integrating v(t) and using the boundary conditions s(1)=2

$s(t) = \int {v(t)} \, dt\\ s(t) = \int {t^3} \, dt\\s(t) = \dfrac{t^4}{4}+c$

we know throught s(1) = 2, that at t=1, s =2. we can use this to find the value of the constant c.

$s(1) = \dfrac{1^4}{4}+c\\4 = \dfrac{1^4}{4}+c\\c = 4-\dfrac{1}{4}\\c = \dfrac{15}{4} = 3.75$

Now we can use this value of t to formulate the position function s(t):

$s(t) = \dfrac{t^4}{4}+\dfrac{15}{4}\\$

this is the position at time t.

to find the position at t=2

$s(2) = \dfrac{2^4}{4}+\dfrac{15}{4}\\$

$s(2) = \dfrac{2^4}{4}+\dfrac{15}{4}\\$

$s(2) = \dfrac{31}{4} = 7.75$

the position of the particle at time, t =2 is s(2) = 7.75

3 0

3 years ago

Cos10+cos80/sin10+sin80=1

eduard

Evaluate <span><span>cos<span>(10)</span></span><span>cos10</span></span> to get <span>0.984807750.98480775</span>.<span><span><span>0.98480775<span>cos<span>(80)</span></span></span><span><span>−<span>sin<span>(10)</span></span></span><span>sin<span>(80)</span></span></span></span><span><span>0.98480775⁢<span>cos80</span></span><span><span>-⁢<span>sin10</span></span>⁢<span>sin80</span></span></span></span>Evaluate <span><span>cos<span>(80)</span></span><span>cos80</span></span> to get <span>0.173648170.17364817</span>.<span><span><span>0.98480775⋅0.17364817</span><span><span>−<span>sin<span>(10)</span></span></span><span>sin<span>(80)</span></span></span></span><span><span>0.98480775⋅0.17364817</span><span><span>-⁢<span>sin10</span></span>⁢<span>sin80</span></span></span></span>Multiply <span>0.984807750.98480775</span> by <span>0.173648170.17364817</span> to get <span>0.171010070.17101007</span>.<span><span>0.17101007<span><span>−<span>sin<span>(10)</span></span></span><span>sin<span>(80)</span></span></span></span><span>0.17101007<span><span>-⁢<span>sin10</span></span>⁢<span>sin80</span></span></span></span>Evaluate <span><span>sin<span>(10)</span></span><span>sin10</span></span> to get <span>0.173648170.17364817</span>.<span><span>0.17101007<span><span><span>−1</span>⋅0.17364817</span><span>sin<span>(80)</span></span></span></span><span>0.17101007<span><span><span>-1</span>⋅0.17364817</span>⁢<span>sin80</span></span></span></span>Multiply <span><span>−1</span><span>-1</span></span> by <span>0.173648170.17364817</span> to get <span><span>−0.17364817</span><span>-0.17364817</span></span>.<span><span>0.17101007<span><span>−0.17364817</span><span>sin<span>(80)</span></span></span></span><span>0.17101007<span><span>-0.17364817</span>⁢<span>sin80</span></span></span></span>Evaluate <span><span>sin<span>(80)</span></span><span>sin80</span></span> to get <span>0.984807750.98480775</span>.<span><span>0.17101007<span><span>−0.17364817</span>⋅0.98480775</span></span><span>0.17101007<span><span>-0.17364817</span>⋅0.98480775</span></span></span>Multiply <span><span>−0.17364817</span><span>-0.17364817</span></span> by <span>0.984807750.98480775</span> to get <span><span>−0.17101007</span><span>-0.17101007</span></span>.<span><span>0.17101007<span>−0.17101007</span></span><span>0.17101007<span>-0.17101007</span></span></span>Subtract <span>0.171010070.17101007</span> from <span>0.171010070.17101007</span> to get <span>0</span>.0

7 0

3 years ago