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3 years ago

Find the interquartile range of 17, 23, 8, 5, 9, 16, 22, 11, 13, 15, 17, 18

Mathematics

2 answers:

zzz [600]3 years ago

5 0

The answer is 13 because 22-9 equals 13.

Yuri [45]3 years ago

4 0

8.25

Hope I helped! ( Smiles )

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2 divion 18 I need helpppppppp

Gnoma [55]

1 divided by 18 is .111111111

3 0

3 years ago

Read 2 more answers

If mÃc=170° Find mLB

kenny6666 [7]

Answer:

∠ B = 85°

Step-by-step explanation:

the inscribed angle B is half the measure of its intercepted arc AC , then

∠ B = $\frac{1}{2}$ × 170° = 85°

8 0

2 years ago

35 solve the following syste

blondinia [14]

Part A: The solution is $(-0.923,5.692)$

Part B: The point $(3,7)$ is not in the solution set.

Explanation:

Part A: The given inequalities are $3 x+4 y>20$ and $x$

The solution can be determined by solving the two inequalities by substitution method.

Changing inequalities to equality, we have,

$x=3 y-18$ and $3 x+4 y=20$

Let us substitute $x=3 y-18$ in the equation $3 x+4 y=20$ , we get,

$3 (3y-18)+4 y=20$

$9y-54+4y=20$

$13y=74$

$y=5.692$

Substituting $y=5.692$ in $x=3 y-18$ , we get,

$x=3 (5.692)-18$

$=17.076-18$

$x=-0.923$

Thus, the solution set is $(-0.923,5.692)$

Part B: Now, we shall determine whether the point $(3,7)$ is in the solution set.

Let us substitute the point $(3,7)$ in the inequalities $3 x+4 y>20$ and $x$ , we get,

$3 (3)+4 (7)>20$

$9+28>20$

$37>20$

Also, substituting $(3,7)$ in $x$ , we get,

$3$

Since, the point $(3,7)$ does not satisfy one of the inequality $x$ , the solution set does not contain the point $(3,7)$

Thus, the point $(3,7)$ is not in the solution set.

8 0

3 years ago

A bacteria colony increases in size at a rate of 4.0581e1.6t bacteria per hour. If the initial population is 36 bacteria, find t

Harrizon [31]

Answer:

1560

Step-by-step explanation:

The rate of Increase of the population (P) of the bacteria is given as:

$\frac{dP}{dt} =4.058e^{1.6t}$

$dP =4.058e^{1.6t}}dt\\Taking\: Integrals\\\int dP =4.058 \int e^{1.6t}dt\\P(t)=\frac{4.058}{1.6} (e^{1.6t} +K)\\P(t)=2.53625 (e^{1.6t} +K)$

Where k is a constant of Integration.

At t=0, P(t)=36

$36=2.53625 (e^{1.6*0} +K)\\36=2.53625 (e^{0} +K)\\36=2.53625 (1 +K)\\36=2.53625 +2.53625K\\K=13.19$

Therefore:

$P(t)=2.53625 (e^{1.6t} +13.19)\\At \: t=4\\P(4)=2.53625 (e^{1.6*4} +13.19)\\=1559.88\\P(4)=1560$

3 0

3 years ago

You are a waterman daily plying the waters of Chesapeake Bay for blue crabs (Callinectes sapidus), the best-tasting crustacean i

notsponge [240]

The question given is incomplete, I googled and got the complete question as below:

You are a waterman daily plying the waters of Chesapeake Bay for blue crabs (Callinectes sapidus), the best-tasting crustacean in the world. Crab populations and commercial catch rates are highly variable, but the fishery is under constant pressure from over-fishing, habitat destruction, and pollution. These days, you tend to pull crab pots containing an average of 2.4 crabs per pot. Given that you are economically challenged as most commercial fishermen are, and have an expensive boat to pay off, you’re always interested in projecting your income for the day. At the end of one day, you calculate that you’ll need 7 legal-sized crabs in your last pot in order to break even for the day. Use these data to address the following questions. Show your work.

a. What is the probability that your last pot will have the necessary 7 crabs?

b. What is the probability that your last pot will be empty?

Answer:

a. Probability = 0.0083

b. Probability = 0.0907

Step-by-step explanation:

This is Poisson distribution with parameter λ=2.4

The probability that your last pot will have the necessary 7 crabs is calculated below:

P(X=7)= {e-2.4*2.47/7!} = 0.0083

The probability that your last pot will be empty is calculated as:

P(X=0)= {e-2.4*2.40/0!} = 0.0907

8 0

3 years ago