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2 years ago

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12. A scientist is observing a family of mice. There are four nice which are reproducing and doubling in number every month. Whi

ch function best represents the number of mice where x is the number of months that have passed?

1 answer:

Margaret [11]2 years ago

3 0

Answer:

?

x = the number of months pass right

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Find the x-intercepts/roots of the polynomial function. Also, identify the degree of the polynomial

Mrac [35]

Answer:

Step-by-step explanation:

7.

roots=-2,-1,1,3

critical points=(-1.5,-3),(0,6),(2,-13)

absolute minimum=-13

End behavior :approaches +infinity

Relative max=6 at (0,6)

Relative min. =-3 at (-1.5,-3)

and -13 at (2,-13)

interval of increase=(-1.5,0)∪(2,∞)

interval of decrease=(-∞,-1.5)∪(0,4)

8.

roots=-1,2

critical points=(0.5,10.5),(4,-10.5)

Abs.max/abs.min=not defined

End behavior:-∞ to +∞

Relative max=10.5

Relative min=-10.5

interval of increase=(-∞,0.5) ∪ (4,∞)

interval of decrease=(0.5,4)

4 0

3 years ago

If 1/3 of a factory is female and we want to sample workers at this factory, then approximately 1/3 of the sample should be fema

lilavasa [31]

I think the answer is true

6 0

2 years ago

Matthew invested $3,000 into two accounts. One account paid 3% interest and the other paid 8% interest. He earned 4% interest on

boyakko [2]

<u>Answer:</u>

<em>Mathew invested</em><em> $600 and $2400</em><em> in each account.</em>

<u>Solution:</u>

From question, the total amount invested by Mathew is $3000. Let p = $3000.

Mathew has invested the total amount $3000 in two accounts. Let us consider the amount invested in first account as ‘P’

So, the amount invested in second account = 3000 – P

Step 1:

Given that Mathew has paid 3% interest in first account .Let us calculate the simple interest $(I_1)$ earned in first account for one year,

$\text {simple interest}=\frac{\text {pnr}}{100}$

Where

p = amount invested in first account

n = number of years

r = rate of interest

hence, by using above equation we get $(I_1)$ as,

$I_{1}=\frac{P \times 1 \times 3}{100}$ ----- eqn 1

Step 2:

Mathew has paid 8% interest in second account. Let us calculate the simple interest $(I_2)$ earned in second account,

$I_{2} = \frac{(3000-P) \times 1 \times 8}{100} \text { ------ eqn } 2$

Step 3:

Mathew has earned 4% interest on total investment of $3000. Let us calculate the total simple interest (I)

$I = \frac{3000 \times 1 \times 4}{100} ----- eqn 3$

Step 4:

Total simple interest = simple interest on first account + simple interest on second account.

Hence we get,

$I = I_1+ I_2 ---- eqn 4$

By substituting eqn 1 , 2, 3 in eqn 4

$\frac{3000 \times 1 \times 4}{100} = \frac{P \times 1 \times 3}{100} + \frac{(3000-P) \times 1 \times 8}{100}$

$\frac{12000}{100} = \frac{3 P}{100} + \frac{(24000-8 P)}{100}$

12000=3P + 24000 - 8P

5P = 12000

P = 2400

Thus, the value of the variable ‘P’ is 2400

Hence, the amount invested in first account = p = 2400

The amount invested in second account = 3000 – p = 3000 – 2400 = 600

Hence, Mathew invested $600 and $2400 in each account.

3 0

3 years ago

Read 2 more answers

Which digit has the greatest value in the number 1,567

Nookie1986 [14]

Seven has the greatest value as a digit, but 1, 000 has the greatest standard form value.

7 0

3 years ago

Constance invested $4000 for 3 years in a savings account paying simple interest with a yearly interest rate of 3.5%. How much s

castortr0y [4]

To solve this problem you might have to use this formula y=a(1+r)^t A=amount of money you already have 1= a set number R= rate T= time YOUR GIVENS ARE: A=4000 R= 3.5% (0.035) decimals T= 3 years NOW lets plug IN everything in the FORMULA y=4000(1+.035)^3= $4434.87. NOW lets subtract 4000-4434=434. She earned $434 of simple interest.

7 0

3 years ago

Read 2 more answers