Ask question

Ask question

All categories

3 years ago

11

A music producer is making a list of vocalists needed to record an album. For each day of recording, a different number of vocal

ists are needed. The first day, eight vocalists are needed. Each day after that, the number of vocalists needed doubles. The producer must pay by the day for each vocalist. To find the total price, the producer needs to know how many vocalists sang in total at the end of the 10th day. Use a series to find the sum after the 10th day.

1 answer:

Angelina_Jolie [31]3 years ago

3 0

Answer:

8184 vocalists sang in total

Step-by-step explanation:

The number needed is ...

8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 + 2048 + 4096

You can add these up to get a total of 8184, or you can use the formula for the sum of a geometric series:

Sn = a1(r^n -1)/(r -1) . . . . where a1 is the first term and r is the common ratio

S10 = 8(2^10 -1)/(2 -1) = 8(1024 -1)/1 = 8184

You might be interested in

(3×1/2) + (6×3/4) + (3×1) + 1 1/4 +1 1/2=<br>

mixer [17]

(3×1/2) + (6×3/4) + (3×1) + 1 1/4 +1 1/2=11.75

8 0

2 years ago

Find the difference (6d+5)−(2−3d)

LiRa [457]

Answer:

9d +3

Step-by-step explanation:

plz brainliestt

5 0

3 years ago

Read 2 more answers

2/(a-7)(a+2)=4/(a+3)(a+2)

pochemuha

$\frac{2}{(a - 7)(a + 2)} = \frac{4}{(a + 3)(a + 2)}$

$\frac{2}{(a^{2} + 2a - 7a - 14)} = \frac{4}{(a^{2} + 2a + 3a + 6)}$

$\frac{2}{a^{2} - 5a - 14} = \frac{4}{a^{2} + 5a + 6}$

$2(a^{2} + 5a + 6) = 4(a^{2} - 5a - 14)$
$2(a^{2}) + 2(5a) + 2(6) = 4(a^{2}) - 4(5a) - 4(14)$
$2a^{2} + 10a + 12 = 4a^{2} - 20a - 56$
$2a^{2} - 2a^{2} + 10a + 12 = 4a^{2} - 2a^{2} - 20a - 56$
$10a + 20a + 12 = 2a^{2} - 20a + 20a - 56$
$30a + 12 + 56= 2a^{2} - 56 + 56$
$\frac{30a + 68}{2} = \frac{2a^{2}}{2}$
$15a + 34 = a^{2}$
$\sqrt{15a + 34} = a$

6 0

3 years ago

Write an equation for the description.<br> Ten times the sum of half a number x and 5 is 9.

Tatiana [17]

This • means Times btw

10•(1/2x+5)=9

8 0

3 years ago

a lunch stand makes a $.75 profit on each chef's salad and $1.20 profit on each caesar salad. On a typical weekday, it sells bet

tangare [24]

Answer:

<em>50 Chef's salads and 50 Caesar salads should be prepared in order to maximize profit.</em>

Step-by-step explanation:

Suppose, the number of Chef's salad is $x$ and the number of Caesar salad is $y$

On a typical weekday, it sells between 40 and 60 Chefs salads and between 35 and 50 Caesar salads.

So, the two constraints are: $40\leq x\leq 60$ and $35\leq y\leq 50$

The total number sold has never exceed 100 salads. So, another constraint will be: $x+y\leq 100$

According to the graph of the constraints, the vertices of the common shaded region are: $(40,35), (60,35), (60,40), (50,50)$ and $(40,50)$ <em>(Refer to the attached image for the graph)</em>

The lunch stand makes a $.75 profit on each Chef's salad and $1.20 profit on each Caesar salad. So, the profit function will be: $P=0.75x+1.20y$

For (40, 35) , $P=0.75(40)+1.20(35)=72$

For (60, 35) , $P=0.75(60)+1.20(35)=87$

For (60, 40) , $P=0.75(60)+1.20(40)=93$

For (50, 50) , $P=0.75(50)+1.20(50)=97.5$ <u><em>(Maximum)</em></u>

For (40, 50) , $P=0.75(40)+1.20(50)=90$

Profit will be maximum when $x=50$ and $y=50$

Thus, 50 Chef's salads and 50 Caesar salads should be prepared in order to maximize profit.

6 0

4 years ago