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Zarrin [17]
3 years ago
6

What are 6 whole numbers -3.5 and 5.5

Mathematics
1 answer:
Blababa [14]3 years ago
3 0

Answer:

This is    a long    long title      restart

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Please help me solve this !:) <br> Explain your answer <br><br> I will give braisnlt . Oh
ser-zykov [4K]

Answer:

D

Step-by-step explanation:

A is incorrect because if 1 x value corresponds to multiple y values, it is not a function

B is incorrect for the same reason as a

C is incorrect because it is linear

D is correct because 1.) all of the others are wrong, and 2.) it is nonlinear.

8 0
3 years ago
Triangle DABC is similar to DEFG.
lyudmila [28]

Answer:

is there an image provided?

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Suppose the average price for new cars in 2012 has a mean of $30,100 and a standard deviation of $5,600. Based on this informati
KatRina [158]

Answer:

($13,300,$46,900)

Step-by-step explanation:

We are given the following in he question:

Mean, μ = $30,100

Standard Deviation, σ = $5,600

Chebyshev's Theorem:

  • According to  theorem atleast 1 - \dfrac{1}{k^2}  percent of data lies within 2 standard deviations of mean.
  • For k = 3,

1 -\dfrac{1}{(3)^2} = 0.889 \approx 89\%

Thus, 89% of data lies within three standard deviation of mean.

\mu - 3(\sigma) = 30100 - 3(5600) = 13300\\\mu + 3(\sigma) = 30100 + 3(5600) = 46900

Thus, we expect at least 89% of new car prices to fall within ($13,300,$46,900)

3 0
3 years ago
You downloaded a video game to your computer. You have a 60-minute free trial of the game. It takes 5 minutes to set up the game
Luba_88 [7]

Answer:

I think the equation is 7l +5=60

Step-by-step explanation:


7 0
4 years ago
Internet company Gurgle is carrying out testing on the efficiency of its search engine. A sample of 31 searches have been carrie
Mrac [35]

Answer:

d. Approximate the standard normal distribution with the Student's t distribution

(0.2199 ; 0.2327)

Step-by-step explanation:

Given that :

Sample size, n = 31

Sample mean, xbar = 0.2258

Sample standard deviation, s = 0.0188

Confidence interval (C. I) :

xbar ± margin of error

Margin of Error : Tcritical * s/sqrt(n)

Degree of freedom, df = n - 1 = 31 - 1 = 30

Tcritical value :

T0.05/2, 30 = 2.042

Margin of Error = 2.042 * 0.0188/sqrt(31)

Margin of Error = 0.0068949

C. I = 0.2258 ± 0.0068949

Lower boundary : (0.2258 - 0.006895) = 0.2189

Upper boundary : (0.2258 - 0.006895) = 0.2327

(0.2199 ; 0.2327)

3 0
3 years ago
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