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ra1l [238]
2 years ago
10

Henry divided his socks into five groups. Let s represent the total number of socks. Which expression and solution represent the

number of socks in each group if s=20?
A. S/5; when s=20, the number of socks in each group is 4.

B. S/5; when s=20, the number of socks in each group is 15.

C. S-5; when s=20, the number of socks in each group is 4.

D. s-5; when s=20, the number of socks in each group is 15.
Mathematics
1 answer:
Triss [41]2 years ago
8 0

Answer:

A. 4 socks per group and S/5

Step-by-step explanation:

20 total # of socks ÷ 5 groups = 4 socks per group

So A. is correct.

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Which equation passes through the points ( -4,1 ) (-8,13 )
Mashutka [201]

Answer:

y= = -3x -11

STANDARD FORM: 3x+y= -11

Step-by-step explanation:

4 0
3 years ago
please can someone help me with this question? i will mark brainliest and i would also really appreciate it! :)
Sloan [31]

Answer:

y ≤ 3x- 1

Step-by-step explanation:

The line is solid, so your answer must have a ≤ or ≥ symbol. And the area below the line is shaded so you would also know you answer must have a ≤ or < symbol.

8 0
2 years ago
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The fox population in a certain region has a continuous growth rate of 5% per year. It is estimated that the population in the y
kvasek [131]

Answer:

P(t) = A * (1 + r)^t ;

14,922 ;

Year 2013

Step-by-step explanation:

Given the following :

Continuous growth rate(r) = 5% = 0.05

Population in year 2000 = Initial population (A) = 10,100

Time(t) = period (years since year 2000)

A)

Find a function that models the population,P(t) , after (t) years since year 2000 (i.e. t= 0 for the year 2000).

P(t) = A * (1 + r)^t

Trying out our function for t = year 2000, t =0

P(0) = 10,100 * (1 + 0.05)^0

P(0) = 10,100 * 1.05^0 = 10,100

B.)

Use your function from part (a) to estimate the fox population in the year 2008.

Year 2008, t = 8

P(8) = 10,100 * (1 + 0.05)^8

P(8) = 10,100 * 1. 05^8

P(8) = 10,100 * 1.4774554437890625

= 14922.29

= 14,922

c) Use your function to estimate the year when the fox population will reach over 18,400 foxes. Round t to the nearest whole year, then state the year.

P(t) = A * (1 + r)^t

18400 = 10,100 * (1.05)^t

18400/10100 = 1.05^t

1.8217821 = 1.05^t

1.05^t = 1.8217821

In(1.05^t) = ln(1.8217821)

0.0487901 * t = 0.5998151

t = 0.5998151 / 0.0487901

t = 12.293787

Therefore eit will take 13 years

2000 + 13 = 2013

4 0
3 years ago
The random variable X is exponentially distributed, where X represents the waiting time to be seated at a restaurant during the
erastova [34]

Answer:

The probability that the wait time is greater than 14 minutes  is 0.4786.

Step-by-step explanation:

The random variable <em>X</em> is defined as the waiting time to be seated at a restaurant during the evening.

The average waiting time is, <em>β</em> = 19 minutes.

The random variable <em>X</em> follows an Exponential distribution with parameter \lambda=\frac{1}{\beta}=\frac{1}{19}.

The probability distribution function of <em>X</em> is:

f(x)=\lambda e^{-\lambda x};\ x=0,1,2,3...

Compute the value of the event (<em>X</em> > 14) as follows:

P(X>14)=\int\limits^{\infty}_{14} {\lambda e^{-\lambda x}} \, dx=\lambda \int\limits^{\infty}_{14} {e^{-\lambda x}} \, dx\\=\lambda |\frac{e^{-\lambda x}}{-\lambda}|^{\infty}_{14}=e^{-\frac{1}{19} \times14}-0\\=0.4786

Thus, the probability that the wait time is greater than 14 minutes  is 0.4786.

7 0
3 years ago
HELP ASAP
tester [92]

Answer:

yolo

Step-by-step explanation:

5 0
3 years ago
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