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monitta
3 years ago
6

What is twenty devided by 5 equalls what?

Mathematics
1 answer:
Alona [7]3 years ago
8 0
20 divided by 5 equals 4

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Find the slope and y-intrcept of <br><br>y=5/3x -4 ​
natima [27]

Answer:

Slope: 5/3

y-intercept: (0, -4)

Step-by-step explanation:

Slope-Intercept Formula: y = mx + b

<em>y</em> is your output.

<em>m</em> is your slope.

<em>b</em> is your y-intercept.

7 0
3 years ago
Arrange the digits 1,2,3,4,5,6,7 so that when the numbers are added up, they will total 100.
GuDViN [60]
36 + 57  + 4 + 2 + 1 =   100   would do it.
6 0
3 years ago
How do you find the area of a parallelogram
yKpoI14uk [10]
A=B H 

area equals the base times the height 


4 0
4 years ago
A simple random sample of size n is drawn from a population that is normally distributed. The sample​ mean, x overbar​, is found
joja [24]

Answer:

(a) 80% confidence interval for the population mean is [109.24 , 116.76].

(b) 80% confidence interval for the population mean is [109.86 , 116.14].

(c) 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed.

Step-by-step explanation:

We are given that a simple random sample of size n is drawn from a population that is normally distributed.

The sample​ mean is found to be 113​ and the sample standard​ deviation is found to be 10.

(a) The sample size given is n = 13.

Firstly, the pivotal quantity for 80% confidence interval for the population mean is given by;

                               P.Q. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean = 113

             s = sample standard​ deviation = 10

             n = sample size = 13

             \mu = population mean

<em>Here for constructing 80% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.</em>

<u>So, 80% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-1.356 < t_1_2 < 1.356) = 0.80  {As the critical value of t at 12 degree

                                          of freedom are -1.356 & 1.356 with P = 10%}  

P(-1.356 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.356) = 0.80

P( -1.356 \times }{\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 1.356 \times }{\frac{s}{\sqrt{n} } } ) = 0.80

P( \bar X-1.356 \times }{\frac{s}{\sqrt{n} } } < \mu < \bar X+1.356 \times }{\frac{s}{\sqrt{n} } } ) = 0.80

<u>80% confidence interval for </u>\mu = [ \bar X-1.356 \times }{\frac{s}{\sqrt{n} } } , \bar X+1.356 \times }{\frac{s}{\sqrt{n} } }]

                                           = [ 113-1.356 \times }{\frac{10}{\sqrt{13} } } , 113+1.356 \times }{\frac{10}{\sqrt{13} } } ]

                                           = [109.24 , 116.76]

Therefore, 80% confidence interval for the population mean is [109.24 , 116.76].

(b) Now, the sample size has been changed to 18, i.e; n = 18.

So, the critical values of t at 17 degree of freedom would now be -1.333 & 1.333 with P = 10%.

<u>80% confidence interval for </u>\mu = [ \bar X-1.333 \times }{\frac{s}{\sqrt{n} } } , \bar X+1.333 \times }{\frac{s}{\sqrt{n} } }]

                                              = [ 113-1.333 \times }{\frac{10}{\sqrt{18} } } , 113+1.333 \times }{\frac{10}{\sqrt{18} } } ]

                                               = [109.86 , 116.14]

Therefore, 80% confidence interval for the population mean is [109.86 , 116.14].

(c) Now, we have to construct 98% confidence interval with sample size, n = 13.

So, the critical values of t at 12 degree of freedom would now be -2.681 & 2.681 with P = 1%.

<u>98% confidence interval for </u>\mu = [ \bar X-2.681 \times }{\frac{s}{\sqrt{n} } } , \bar X+2.681 \times }{\frac{s}{\sqrt{n} } }]

                                              = [ 113-2.681 \times }{\frac{10}{\sqrt{13} } } , 113+2.681 \times }{\frac{10}{\sqrt{13} } } ]

                                               = [105.56 , 120.44]

Therefore, 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed because t test statistics is used only when the data follows normal distribution.

6 0
3 years ago
What is the volume of the prism?
jeka57 [31]
The volume prism refers to the number of cubic units that will exactly fill the figure. The volume of a rectangular prism can be found or calculate by using the formula V=Bh, where B represents to the area of the base or in other words the length and width of the rectangle.

In this exercise is given that the measurements of a prism are 5/2ft, 3/2ft, and 7/2ft; and it is asked to find its volume. In order to find the volume of the prism, you should substitute the given values into the previous mention formula.

V=Bh
V=(5/2 ft)(7/2 ft)(3/2 ft)
V=(35/4 ft²)(3/2 ft)
V=105/8ft³  or 13 \frac{1}{8}ft³

The volume of the rectangular prism is 13 \frac{1}{8}ft³.

5 0
3 years ago
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