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weqwewe [10]
2 years ago
15

Please answer this. I need and will give

Mathematics
1 answer:
aev [14]2 years ago
3 0
At least give a clear picture bro
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On a calendar of 31 days what are the prime numbers
zepelin [54]
I believe <span>2   3   5   7   11   13   17   19   23   29   31</span>
5 0
3 years ago
Read 2 more answers
PLEASE HELP ASAP!!!
xeze [42]

Answer:

JM = 10

Step-by-step explanation:

Because MN║JL,

\frac{MK}{JM} = \frac{KN}{NL}

\frac{5}{2x + 4}  = \frac{6}{12}  =  \frac{1}{2}

2(5) = 1(2x + 4)

10 = 2x + 4

2x = 6

x = 3

JM = 2x + 4 = 2(3) + 4 = 6 + 4 = 10

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3 years ago
Find the total number of outcomes from rolling a number cube with sides labeled 1-6 and choosing a letter from the word "NUMBERS
Irina18 [472]

Answer:

........................................................................................

Step-by-step explanation:

4 0
3 years ago
Which expression below is equivalent to -3x+5(x+2)
kari74 [83]

Answer:


                                         Step-by-step explanation:

1.) Apply the distributive law

                5x + 5 * 2

2.) Multiply the numbers

       5 * 2 = 10 is equivalent to 5x + 10

3.) Group like terms

                 5x - 3 + 10

4.) Add/subtract the numbers

                  -3 + 10 = 7 is equivalent to 5x + 7




4 0
3 years ago
Suppose that 15% of the fields in a given agricultural area are infested with the sweet potato whitefly. One hundred fields in t
hram777 [196]

Answer:

a.\mu=15

b.\mu=7.8586 \ and  \ \mu=22.1414

c. Choice A- Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.

Step-by-step explanation:

a.Binomial distribution is defined by the expression

P(X=k)=C_k^n.p^k.(1-p)^{n-k}

Let n be the number of trials,n=100

and p be the probability of success,p=15\%

The mean of a binomial distribution is the probability x sample size.

\mu=np=100\times0.15=15

b.Limits within which p is approximately 95%

sd of a binomial distribution is given as:\sigma=\sqrt npq\\q=1-p

Therefore, \sigma=\sqrt(100\times0.015\times0.85)=3.5707

Use the empirical rule to find the limits. From the rule, approximately 95% of the observations are within to standard deviations from mean.

sd_1=>\mu-2\sigma=15-2\times3.3507=7.8586\\sd_2=>\mu-2\sigma=15+2\times3.3507=22.1414

Hence, approximately 95% of the observations are within 7.8586 and 22.1414 (areas of infestation).

c.  x=45 is not within the limits in b above (7.8586,22.1414). X=45 appears to be a large area of infestation. A.Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.

4 0
3 years ago
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