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vredina [299]
2 years ago
11

Suppose that quiz scores in a beginning statistics class have a mean of 6.4 with a standard deviation of 0.6. Using Chebyshev's

Theorem, state the range in which at
least 75% of the data will reside. Please do not round your answers.
Answer
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Mathematics
1 answer:
QveST [7]2 years ago
4 0

(5.2,7.6)

mean=6.4

standard deviation=0.6

we have to find the range in which at least 75% of the data will reside

1-1/k²=0.75

1/k²=1-0.75

1/k²=0.25

k²=1/0.25=4

k=2

so, k=2

The range of values can be computed as mean±k(standard deviation).

Thus, the range in which at least 75% of the data will reside is

(mean-k(standard deviation), mean+ k(standard deviation))

(6.4-2(0.6),6.4+2(0.6)

(6.4-1.2,6.4+1.2)

(5.2,7.6)

learn more of standard deviation here brainly.com/question/14671301

#SPJ9

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which inequality represents all values of x for which the quotient below is defined? ✓15(x-1) ÷ ✓2x^2​
eduard

Answer:

C. x ≥ 1

Step-by-step explanation:

A P E X

6 0
4 years ago
On a farm the number of cows and the number of sheep are in the tatio 6:5 the number of sheep and the number of pigs are in the
Rzqust [24]

Answer:

There are 70 sheep in the farm.

Step-by-step explanation:

Consider the provided information.

Let c represents the cows, s represents the sheep and p represents the pigs.

The number of cows and the number of sheep are in the ratio 6:5.

\dfrac{c}{s} =\dfrac{6}{5}

c =\dfrac{6s}{5}

The number of sheep and the number of pigs are in the ratio 2:1.

\dfrac{s}{p} =\dfrac{2}{1}

p =\dfrac{s}{2}

The total number is 189.

c+s+p=189

Substitute the value of c and p in above.

\dfrac{6s}{5}+s+\dfrac{s}{2}=189

\dfrac{12s+10s+5s}{10}=189

27s=1890

s=70

Hence, there are 70 sheep in the farm.

8 0
3 years ago
Help pls hello pls hello hello
ipn [44]
Basically, we end up with a false statement like
0=3
or something like that
so simplfy each

A.
-5+8x-9=3(x+3)
8x-14=3x+9
minus 3x both sides
5x-14=9
add 15
5x=23
we see we will get a solution

B.
distribute
-12+6x=-12+6x
we can see there are infinite solutions

C.
6-2(3-2x)=-4(3-x)
6-6+4x=-12+4x
0+4x=-12+4x
minus 4x both sides
0=-12
false
here is our answer

C is the one with no solution
 
5 0
3 years ago
Read 2 more answers
How would I solve this? I tried and I got a reallly crazy number... I think I messed up somewhere. The question is... Jack and M
Lera25 [3.4K]
37 x 10 = 370
21 x 11 = 231

Mark worked 21 Hours, Jack worked 37 Hours.
4 0
3 years ago
Read 2 more answers
The amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.0 minutes and
strojnjashka [21]

Answer:

P ( 5 < X < 10 ) = 1

Step-by-step explanation:

Given:-

- Sample size n = 49

- The sample mean u = 8.0 mins

- The sample standard deviation s = 1.3 mins

Find:-

Find the probability that the average time waiting in line for these customers is between 5 and 10 minutes.

Solution:-

- We will assume that the random variable follows a normal distribution with, then its given that the sample also exhibits normality. The population distribution can be expressed as:

                                   X ~ N ( u , s /√n )

Where

                            s /√n = 1.3 / √49 = 0.2143

- The required probability is P ( 5 < X < 10 ) minutes. The standardized values are:

                        P ( 5 < X < 10 ) = P (    (5 - 8) / 0.2143 <  Z  <  (10-8) / 0.2143   )

                                                 = P ( -14.93 < Z < 8.4 )

- Using standard Z-table we have:

                        P ( 5 < X < 10 ) = P ( -14.93 < Z < 8.4 ) = 1        

7 0
3 years ago
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