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2 years ago

The unit rate for $124 for 8 skirts is $

Mathematics

1 answer:

dedylja [7]2 years ago

5 0

Answer:

15.20,15.50

Step-by-step explanation:

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A community college has 2,000 students and 50 instructors. Next year the enrollment will be 6,000. How many new instructors shou

juin [17]

$\begin{array}{ccc}2,000&-&50\\\\6,000&-&x\end{array}\\\\\\2,000x=50\times6,000\ \ \ \ \ /:2,000\\\\x=\frac{300,000}{2,000}\\\\x=150\\\\150-50=100\\\\Answer:100\ new\ instructors.$

7 0

3 years ago

During the holiday sale at blooming department store all sweaters are discounted 30% on blockbuster Saturday the store discounts

RSB [31]

Answer: He would pay $62.37 for it on blockbuster Saturday.

Step-by-step explanation:

Since we have given that

Original price of sweater = $99

Discount on blockbuster Saturday = 30%

Additional discount on sweater = 10%

So, According to question,

$Net\ Price=Price\times \frac{100-Discount}{100}\\\\\text{As there are 2 successive discounts }\\\\So,\\\\Net\ Price=Price\times \frac{100-D_1}{100}\times\frac{100-D_2}{100}$

$99\times \frac{100-30}{100}\times \frac{100-10}{100}\\\\=99\times \frac{70}{100}\times \frac{90}{100}\\\\=\$62.37$

Hence, he would pay $62.37 for it on blockbuster Saturday.

3 0

2 years ago

For the following amount at the given interest rate compounded continuously, find (a) the future value after 9 years, (b) the

Stella [2.4K]

Answer:

(a) The future value after 9 years is $7142.49.

(b) The effective rate is $r_E=4.759 \:{\%}$ .

(c) The time to reach $13,000 is 21.88 years.

Step-by-step explanation:

The definition of Continuous Compounding is

If a deposit of $P$ dollars is invested at a rate of interest $r$ compounded continuously for $t$ years, the compound amount is

$A=Pe^{rt}$

(a) From the information given

$P=4700$

$r=4.65\%=\frac{4.65}{100} =0.0465$

$t=9 \:years$

Applying the above formula we get that

$A=4700e^{0.0465\cdot 9}\\A=7142.49$

The future value after 9 years is $7142.49.

(b) The effective rate is given by

$r_E=e^r-1$

Therefore,

$r_E=e^{0.0465}-1=0.04759\\r_E=4.759 \:{\%}$

(c) To find the time to reach $13,000, we must solve the equation

$13000=4700e^{0.0465\cdot t}$

$4700e^{0.0465t}=13000\\\\\frac{4700e^{0.0465t}}{4700}=\frac{13000}{4700}\\\\e^{0.0465t}=\frac{130}{47}\\\\\ln \left(e^{0.0465t}\right)=\ln \left(\frac{130}{47}\right)\\\\0.0465t\ln \left(e\right)=\ln \left(\frac{130}{47}\right)\\\\0.0465t=\ln \left(\frac{130}{47}\right)\\\\t=\frac{\ln \left(\frac{130}{47}\right)}{0.0465}\approx21.88$