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skelet666 [1.2K]
3 years ago
6

Plz help me giving brainliest

Mathematics
1 answer:
lianna [129]3 years ago
7 0

Answer:

A. 80

Step-by-step explanation:

Count the first top half and you get 40 and you will know the bottom half is 40

which altogether is 80 or you can count the blocks one by one or count the blocks on the right side count the top blocks by five across.

V= l*w*h

V= 5*4*4

V= 80

Hope this Helps! :)

Have a nice day! :)

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What is the greatest common factor of 32 and 48
Bond [772]
Turn each number into the product of it's prime factors.

32=16x2=2x2x2x2x2=2^5

48=24*2=6x4x2=2x3x2x2x2

Pick the highest number that occurs. In this case it is 2. Now we have to see how many times it appears in both. It appears 5 times in 32 and 4 times in 48. 4 is the highest number of times it appears in the numbers so:

2^4=2x2x2x2=16

The Greatest Common Factor (GCF) of 32 and 48 is 16.
4 0
3 years ago
Read 2 more answers
What is 1/2(4x+14)=2(x-7)
GREYUIT [131]

Answer:

no solution

Step-by-step explanation:

1/2(4x+14)=2(x-7)

Multiply both sides by 2.

4x + 14 = 4(x - 7)

4x + 14 = 4x - 28

Subtract 4x from both sides.

14 = -28

Since 14 = -28 is a false statement, there is no solution for this equation.

Answer: no solution

8 0
3 years ago
Need help number 2 for a quiz !!!
Svetach [21]

Answer: radius: 6, center: (3,-4)

Step-by-step explanation:

5 0
3 years ago
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Which of the following is not a property of a chi-square distribution?
laiz [17]

Answer:

c) Is not a property (hence (d) is not either)

Step-by-step explanation:

Remember that the chi square distribution with k degrees of freedom has this formula

\chi_k^2 = \matchal{N}_1^2 +  \matchal{N}_2^2 + ... + \, \matchal{N}_{k-1}^2 +  \matchal{N}_k^2

Where N₁ , N₂m .... N_k are independent random variables with standard normal distribution. Since it is a sum of squares, then the chi square distribution cant take negative values, thus (c) is not true as property. Therefore, (d) cant be true either.

Since the chi square is a sum of squares of a symmetrical random variable, it is skewed to the right (values with big absolute value, either positive or negative, will represent a big weight for the graph that is not compensated with values near 0). This shows that (a) is true

The more degrees of freedom the chi square has, the less skewed to the right it is, up to the point of being almost symmetrical for high values of k. In fact, the Central Limit Theorem states that a chi sqare with n degrees of freedom, with n big, will have a distribution approximate to a Normal distribution, therefore, it is not very skewed for high values of n. As a conclusion, the shape of the distribution changes when the degrees of freedom increase, because the distribution is more symmetrical the higher the degrees of freedom are. Thus, (b) is true.

6 0
3 years ago
Find an equation in standard form for the hyperbola with vertices at (0, ±9) and foci at (0, ±11)
Katen [24]

Answer:

\frac{y^{2} }{81} -\frac{x^{2} }{40} }=1 is standard equation of hyperbola with vertices at (0, ±9) and foci at (0, ±11).

Step-by-step explanation:

We have given the vertices at (0, ±9) and foci at (0, ±11).

Let (0,±a)  = (0,±9) and (0,±c)  = (0,±11)

The standard equation of parabola is:

\frac{y^{2} }{a^{2} } -\frac{x^{2} }{b^{2} }=1

From statement, a  = 9

c² = a²+b²

(11)²  = (9)²+b²

121-81  = b²

40  = b²

Putting the value of a² and b² in standard equation of parabola, we have

\frac{y^{2} }{81} -\frac{x^{2} }{40} }=1 which is the answer.

3 0
3 years ago
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