Can someone please explain this step by step.

Lostsunrise [7]

Answer:

x=2,y=0

(You did not provide enough information for me to know what to do with said equations, so I'm assuming it was System of Equations.)

7 0

2 years ago

Use the following information for problems 1 – 3. Suppose you sign a contract for an annual salary of $50,000 with a guaranteed

erik [133]

Initial salary = $50,000 .

Rate of raise = 5% each year.

Therefore, each next year salary would be 105% that is 1.05 times.

5% of 50,000 = 0.05 × 50000 = 2500.

Therefore raise is $2500 each year.

According to geometric sequence first term 50000 and common ratio 1.05.

Applying geometric sequence formula

$a_n = ar^{n-1}$

1) $a_n = 50000(1.05)^{n-1}$

2) In order to find salary in 5 years we need to plug n=5, we get

$a_5 = 50000(1.05)^{5-1}= 50000(1.05)^4$

= 50000(1.21550625)

3) In order to find the total salary in 10 years we need to apply sum of 10 terms formula of a geometric sequence.

$S_n = \frac{a(1-r^n}{1-r}$

Plugging n=10, a = 50000 and r= 1.05.

$S_10 = \frac{50000(1-(1.05)^{10}}{1-1.05}$

$S_10 = \frac{50000(0.050)^{10}}{0.05}$

= 628894.62678.

<h3>Therefore , you will have earned $ 628894.62678 in total salary by the end of your 10th year.</h3>

7 0

2 years ago

Can someone do this for me please???? i need it done fast!!

KIM [24]

Depends on what it is

6 0

3 years ago

Find the equation of the line shown.<br>

trasher [3.6K]

HOPE IT HELPS!!!!!!!!

4 0

3 years ago

a. Fill in the midpoint of each class in the column provided. b. Enter the midpoints in L1 and the frequencies in L2, and use 1-

Tresset [83]

Answer:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795

Step-by-step explanation:

Given

See attachment for class

Solving (a): Fill the midpoint of each class.

Midpoint (M) is calculated as:

$M = \frac{1}{2}(Lower + Upper)$

Where

$Lower \to$ Lower class interval

$Upper \to$ Upper class interval

So, we have:

Class 63-65:

$M = \frac{1}{2}(63 + 65) = 64$

Class 66 - 68:

$M = \frac{1}{2}(66 + 68) = 67$

When the computation is completed, the frequency distribution will be:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Solving (b): Mean and standard deviation using 1-VarStats

Using 1-VarStats, the solution is:

$\bar x = 70.2903$

$\sigma = 3.5795$

<em>See attachment for result of 1-VarStats</em>

8 0

3 years ago