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konstantin123 [22]
3 years ago
8

I need help as soon as possible

Mathematics
1 answer:
bixtya [17]3 years ago
7 0
I believe the correct answer is -2
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There were a total of 209 basketball games in the season the season played is for 19 months how many basketball games were playe
dangina [55]

Answer:

11

Step-by-step explanation:

divide 209 by 19 to get an equal number of games each month (for 19 months), you will get 11 games each month.

6 0
2 years ago
Read 2 more answers
each truck in the line weigh two tons more than the truck before it. the trucks weigh a total of 32 tons.how many pounds does a
mixer [17]
If there are 4 trucks then the first truck weighs x
then x+2tons
then x+4tons
then x+6tons = 32tons
so 4x+12=32
         -12   -12
     /4         /4
x=5 
so plug it back in 
truck 1= 5tons or 10,000lbs
truck 2= 7tons or 14,000lbs
truck 3= 9tons or 18,000lbs
<span>truck 4= 11tons or 22,000lbs</span>
6 0
3 years ago
Compare the rates to find which is the best value.
sattari [20]

Answer:

C6 greeting cards for $23.40

Step-by-step explanation:

A. 13.5/3=4.5

B. 38.25/9=4.25

C. 23.4/6=3.9

d. 32.8/8=4.1

5 0
2 years ago
In a population of similar households, suppose the weekly supermarket expense for a typical household is normally distributed wi
Rina8888 [55]

Answer:

P(Y ≥ 15) = 0.763

Step-by-step explanation:

Given that:

Mean =135

standard deviation = 12

sample size n  = 50

sample mean \overline x = 140

Suppose X is the random variable that follows a normal distribution which represents the weekly supermarket expenses

Then,

X \sim N ( \mu \sigma)

The probability that X is greater than 140 is :

P(X>140) = 1 - P(X ≤ 140)

P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{140-135}{12})

P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{5}{12})

P(X>140) = 1 - P( Z\leq0.42)

From z tables,

P(X>140) = 1 - 0.6628

P(X>140) = 0.3372

Similarly, let consider Y to be the variable that follows a binomial distribution of the no of household whose expense is greater than $140

Then;

Y \sim Binomial (np)

Y \sim Binomial (50,0.3372)

∴

P(Y ≥ 15) = 1- P(Y< 15)

P(Y ≥ 15) = 1 - ( P(Y=0) + P(Y=1) + P(Y=2) + ... + P(Y=14) )

P(Y \geq 15) = 1 - \begin {pmatrix} ^{50}_0 \end {pmatrix} (0.3372)^0 (1-0,3372)^{50} + \begin {pmatrix} ^{50}_1 \end {pmatrix} (0.3372)^1 (1-0,3372)^{49}  + \begin {pmatrix} ^{50}_2 \end {pmatrix} (0.3372)^2 (1-0,3372)^{48} +...  + \begin {pmatrix} ^{50}_{50{ \end {pmatrix} (0.3372)^{50} (1-0,3372)^{0}

P(Y ≥ 15) = 0.763

7 0
3 years ago
In a semiconductor company's quality control test, a machine found that 14 out of a sample of 234 computer chips were defective.
ivann1987 [24]

Answer:

281-282

Step-by-step explanation:

4 0
2 years ago
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