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sasho [114]
3 years ago
15

Find the median and mean of the data set below:

Mathematics
2 answers:
EastWind [94]3 years ago
7 0

Answer:

2 11 33 41 44 49

Step-by-step explanation:

33+41=74 74÷2=37

KengaRu [80]3 years ago
6 0

Answer: The medians the middle, so you put in order to 2, 11, 33, 41, 44, and 49. Then, you just pick the number in the middle and since it's 33 and 41, you add them together then divide by 2 and you answer is 37. I hope this helps you!

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What exponent would makethis statment true 100,000=10?​
Phantasy [73]

Answer:

5

Step-by-step explanation:

Fifth-grade math class taught me.

7 0
3 years ago
Write the product in its simplest form: -5x*8x^7
Ivenika [448]

Answer:

-40 x^8

Step-by-step explanation:

-5x*8x^7

We know that a^b * a^c = a^(b+c)

-5 *8  * (x*x^7)

-40  *(x^1 * x^7)

-40  x^(1+7)

-40 x^8

4 0
4 years ago
Read 2 more answers
104 pounds requires 156 milligrams of medicine what is the weight of a patient who requires 207 milligrams of medicine
Lubov Fominskaja [6]

Answer:

138 pounds

Step-by-step explanation:

Let the required weight of the patient be x pounds.

We have that,

156 milligrams of medicine is required for a patient weighing 104 pounds.

Then, 207 milligrams of medicine will be required for a patient having weight as shown below,

Weight of the patient = \frac{104 \times 207}{156}

i.e. x = \frac{21528}{156}

i.e. x = 138

Hence, the patient needing 207 milligrams of medicine will have the weight 138 pounds.

5 0
3 years ago
UM PLZ HELP MEEEEEEE
raketka [301]
The answer is A. The first one
8 0
3 years ago
Read 2 more answers
Goranson and Hall (1980) explain that the probability of detecting a crack in an airplane wing is the product of p1, the probabi
PolarNik [594]

Answer:

a. The multiplication of these probabilities is justified because the inspections are isolated or don't have the same goal. For instance p1, would mean that not all the planes are checked, p2 that not all the parts in the plane are checked, and p3, that even if the the part where the crack could be inspected, people checking it could not notice it or it could be not identified easily.

b. As the events are independent and there are only two possible answers (detect the plane or not), a binomial distribution could be applied, therefore:

p=0.9*0.8*.5=36 (probability of detecting the crack)

n=3 (number of possibilities, in this case number of planes)

The probability in a binomial formula is given by

P(X=x)=\frac{n!}{(n-x)!x!}*p^{x}*(1-p)^{n-x}

Considering that the only possibility that is not being asked is the one of not detecting any crack which would mean x=0, then, we could find the probability as

P(X\geq 1)=1-P(X=0)

P(X\geq 1)=1-P(X=0)=1-\frac{3!}{3!0!} *0.36^{0} *0.64^{3} =0.737856

Probability is then 0.737856

Step-by-step explanation:

5 0
3 years ago
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