1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Fittoniya [83]
3 years ago
10

Spencer has already taken 4 quizzes during past quarters, and he expects to have 5 quizzes during each week of this quarter. How

many weeks of school will Spencer have to attend this quarter before he will have taken a total of 49 quizzes?
_weeks
HELP:')​
Mathematics
1 answer:
babunello [35]3 years ago
6 0

Answer:

<em>spencer will have 9 quizzes</em>

<h3 />

Step-by-step explanation:

You might be interested in
Solve 10x-3=6x+85 give a reason to justify each statement
siniylev [52]
<span>10x-3=6x+85
4x = 88
  x = 22</span>
6 0
3 years ago
Read 2 more answers
Simplify the expression below. 2^3 x 2^2
Zigmanuir [339]

Answer:

32 or~2^5

Step-by-step explanation:

→ (2^3)(2^2)

=2^3*2^2

=(2*2*2)*(2*2)

=2*2*2*2*2

=32

<u><em>You can also put this in another way:</em></u>

<em />2^5=2^3*2^2=2^3^+^2=2^5<em />

7 0
3 years ago
Read 2 more answers
Which of the following is an equivalent ratio for the 15/24? <br> 5/4<br> 30/40<br> 5/8
hoa [83]

Answer:

5/8

Step-by-step explanation:

15/3=5 and 24/3=8

3 0
3 years ago
Read 2 more answers
Question Help Suppose that the lifetimes of light bulbs are approximately normally​ distributed, with a mean of 5656 hours and a
koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean \mu and standard deviation \sigma, the z-score of a value X is given by:

Z = \frac{X - \mu}{\sigma}

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally​ distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So \mu = 5656, \sigma = 333.3

(a) What proportion of light bulbs will last more than 6262 ​hours?

The pvalue of the z-score of X = 6262 is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.

Z = \frac{X - \mu}{\sigma}

Z = \frac{6262 - 5656}{333.3}

Z = 1.82

Z = 1.81 has a pvalue of .96562. This means that 96.562% of the light bulbs are going to last less than 6262 hours. So

P = 100% - 96.562% = 3.438% of the light bulbs will last more than 6262 hours.

​(b) What proportion of light bulbs will last 5252 hours or​ less?

This is the pvalue of the zscore of X = 5252

Z = \frac{X - \mu}{\sigma}

Z = \frac{5252- 5656}{333.3}

Z = -1.21

Z = -1.21 has a pvalue of .1131. This means that 11.31% of the light bulbs will last 5252 hours or less.

(c) What proportion of light bulbs will last between 5858 and 6262 ​hours?

This is the pvalue of the zscore of X = 6262 subtracted by the pvalue of the zscore X = 5858

For X = 6262, we have that Z = 1.81 with a pvalue of .96562.

For X = 5858

Z = \frac{X - \mu}{\sigma}

Z = \frac{5858- 5656}{333.3}

Z = 0.61

Z = 0.61 has a pvalue of .72907.

So, the proportion of light bulbs that will last between 5858 and 6262 hours is

P = .96562 - .72907 = .23655

23.655% of the light bulbs are going to last between 5858 and 6262 hours.

​(d) What is the probability that a randomly selected light bulb lasts less than 4646 ​hours?

This is the pvalue of the zscore of X = 4646

Z = \frac{X - \mu}{\sigma}

Z = \frac{4646- 5656}{333.3}

Z = -3.03

Z = -3.03 has a pvalue of .0012. This means that 0.12% of the light bulbs will last 4646 hours or less.

5 0
3 years ago
A company is considering the purchase of a new machine for $75,660. management predicts that the machine can produce sales of $2
nekit [7.7K]

The company is considering the purchase of a new machine for $75,660 (based on the available data), and the payback period is <u>24 years</u>.

<h3>What is the payback period?</h3>

The payback period is the time the company requires to recoup its investment for the new machine.

The payback period can be computed by dividing the investment cash outflows by the annual net cash inflows.

<h3>Data and Calculations:</h3>

Initial investment in new machine = $75,660

Annual depreciation expense = $4,600

Investment period = 10 years

Annual sales revenue = $20,000

Annual expenses = $16,800

Ne annual cash inflow = $3,200 ($20,000 - $16,800)

Payback period = 24 years ($75,660/$3,200)

Thus, since the payback period is <u>24 years</u>, while the investment period is 10 years, it sounds unwise for the company to continue the investment.

Learn more about the payback period at brainly.com/question/23149718

#SPJ1

3 0
1 year ago
Other questions:
  • Write an equation of each line that passes through the following points in slope-intercept form:
    7·1 answer
  • Help with this question!
    6·1 answer
  • Explain how you can use multiplication to find the quotient 3/5 divided by 3/15. Then evaluate the expression
    9·1 answer
  • What is the square root of -1?
    7·1 answer
  • B. suppose ​p(2​)equals​p(3​)equals​p(4​)equals​p(8​)equals​p(9​)equalsstartfraction 1 over 20 endfraction and ​p(1​)equals​p(5​
    12·1 answer
  • Chris sold red and blue pens for a.
    10·2 answers
  • Help with math nation please!!!
    10·1 answer
  • Hellohello, if you able to help me then please do. (:
    11·1 answer
  • Help me pleaseeeeeeeeeeeee
    7·1 answer
  • Simplify the expression:<br><br> 2(3 + g) =
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!