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lara [203]
3 years ago
11

the sum of three consecutive odd integers is less than 54. what is the set of the largest three consecutive odd integers that fi

ts this situation
Mathematics
2 answers:
Mandarinka [93]3 years ago
6 0
The largest of the three would be 53, then 51, then 49.
Ne4ueva [31]3 years ago
3 0
54>15+17+19

15+17+19=51
You might be interested in
When subtracting decimals with the same sign you ______________.
Rus_ich [418]

Hey there!

Answer:\boxed{\text{Line up the decimal}}

Explanation:

When always adding or subtracting, you <em>always</em> have to <u>line up the decimal</u>.

When subtracting decimals with the same sign you <u>Line up the decimal</u>.

Hope this helps!

3 0
3 years ago
Read 2 more answers
Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases: a. Central area 5 .
Flauer [41]

Answer:

a) "=T.INV(0.025,10)" and "=T.INV(1-0.025,10)"

And we got t_{\alpha/2}=-2.228 , t_{1-\alpha/2}=2.228

b)  "=T.INV(0.025,20)" and "=T.INV(1-0.025,20)"

And we got t_{\alpha/2}=-2.086 , t_{1-\alpha/2}=2.086

c) "=T.INV(0.005,20)" and "=T.INV(1-0.005,20)"

And we got t_{\alpha/2}=-2.845 , t_{1-\alpha/2}=2.845

d) "=T.INV(0.005,50)" and "=T.INV(1-0.005,50)"

And we got t_{\alpha/2}=-2.678 , t_{1-\alpha/2}=2.678

e) "=T.INV(1-0.01,25)"

And we got t_{\alpha}= 2.485

f) "=T.INV(0.025,5)"

And we got t_{\alpha}= -2.571

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.  

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

We will use excel in order to find the critical values for this case

Determine the t critical value(s) that will capture the desired t-curve area in each of the following cases:

a. Central area =.95, df = 10

For this case we want 0.95 of the are in the middle so then we have 1-0.95 = 0.05 of the area on the tails. And on each tail we will have \alpha/2=0.025.

We can use the following excel codes:

"=T.INV(0.025,10)" and "=T.INV(1-0.025,10)"

And we got t_{\alpha/2}=-2.228 , t_{1-\alpha/2}=2.228

b. Central area =.95, df = 20

For this case we want 0.95 of the are in the middle so then we have 1-0.95 = 0.05 of the area on the tails. And on each tail we will have \alpha/2=0.025.

We can use the following excel codes:

"=T.INV(0.025,20)" and "=T.INV(1-0.025,20)"

And we got t_{\alpha/2}=-2.086 , t_{1-\alpha/2}=2.086

c. Central area =.99, df = 20

 For this case we want 0.99 of the are in the middle so then we have 1-0.99 = 0.01 of the area on the tails. And on each tail we will have \alpha/2=0.005.

We can use the following excel codes:

"=T.INV(0.005,20)" and "=T.INV(1-0.005,20)"

And we got t_{\alpha/2}=-2.845 , t_{1-\alpha/2}=2.845

d. Central area =.99, df = 50

  For this case we want 0.99 of the are in the middle so then we have 1-0.99 = 0.01 of the area on the tails. And on each tail we will have \alpha/2=0.005.

We can use the following excel codes:

"=T.INV(0.005,50)" and "=T.INV(1-0.005,50)"

And we got t_{\alpha/2}=-2.678 , t_{1-\alpha/2}=2.678

e. Upper-tail area =.01, df = 25

For this case we need on the right tail 0.01 of the area and on the left tail we will have 1-0.01 = 0.99 , that means \alpha =0.01

We can use the following excel code:

"=T.INV(1-0.01,25)"

And we got t_{\alpha}= 2.485

f. Lower-tail area =.025, df = 5

For this case we need on the left tail 0.025 of the area and on the right tail we will have 1-0.025 = 0.975 , that means \alpha =0.025

We can use the following excel code:

"=T.INV(0.025,5)"

And we got t_{\alpha}= -2.571

8 0
3 years ago
Which is a counterexample for the conditional statement shown?
Arte-miy333 [17]

Answer: First Option

<em>The points have the same x-coordinate value.</em>

Step-by-step explanation:

By definition, a relation is considered a function if and only if for each input value x there exists <u><em>only one </em></u>output value y.

So, the only way that the line that connects two points in the coordinate plane is not a function, is that these two points have the same coordinate for x.

For example, suppose you have the points (2, 5) and (2, 8) and draw a line that connects these two points.

The line will be parallel to the y axis.

Note that the value of x is the same x = 2. But when x = 2 then y = 5 and y = 8.

There <u><em>are two output</em></u><em> </em>values (y = 8, y = 5) for the same input value x = 2.

In fact all the vertical lines parallel to the y-axis have infinite output values "y" for a single input value x. Therefore, they can not be defined as a function.

<u>Then the correct option is: </u>

<em>The points have the same x-coordinate value.</em>

7 0
3 years ago
Y=x<br> y=-6x–14<br><br> (substitution method)
olga_2 [115]

Answer:

The solution is the point (-2,-2)

Step-by-step explanation:

we have

y=x ----> equation A

y=-6x-14 ----> equation B

Solve the system by substitution

substitute equation A in equation B

x=-6x-14

Solve for x

6x+x=-14\\7x=-14\\x=-2

Find the value of y

substitute the value of x in the equation A

y=x

so

y=-2

therefore

The solution is the point (-2,-2)

3 0
3 years ago
Find the equation of the ellipse with the following properties:
steposvetlana [31]

Answer: The equation of an ellipse:

(

x

−

h

)

2

a

2

+

(

y

−

k

)

2

b

2

=

1

;

a

>

b

Has vertices at

(

h

±

a

,

k

)

Has foci at

(

h

±

√

a

2

−

b

2

,

k

)

Use the vertices to write 3 equations:

k

=

4

[1]

h

−

a

=

−

6

[2]

h

+

a

=

10

[3]

Use equations [2] and [3] to solve for h and a:

2

h

=

4

h

=

2

a

=

8

Use the focus to write another equation:

8

=

h

+

√

a

2

−

b

2

Substitute values for h and a:

8

=

2

+

√

8

2

−

b

2

6

=

√

64

−

b

2

36

=

64

−

b

2

b

2

=

64

−

36

b

2

=

28

b

=

√

28

Substitute the values into the standard form:

(

x

−

2

)

2

8

2

+

(

y

−

4

)

2

(

√

28

)

2

=

1

5 0
3 years ago
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