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Leviafan [203]
3 years ago
8

Help me pls solve this

Mathematics
2 answers:
Serjik [45]3 years ago
7 0

Answer:

1/8  (Decimal: 0.125)

Step-by-step explanation:


Lina20 [59]3 years ago
4 0

Answer:

⅛ or 0.125

Step-by-step explanation:

5/4 ÷ 10                        Convert division to multiplication by fraction

5/4 ÷ 10 = 5/4 × 1/10     Divide numerator and denominator by 5

5/4 ÷ 10 = 1/4 × ½         Multiply the fractions

5/4 ÷ 10 = ⅛

If you want a decimal fraction,

5/4 ÷ 10 = 0.125


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How to divided 245 by 70 <br> Show your work
Bas_tet [7]

Answer:

Step-by-step explanation:

Hello!

          2          4              5          ∟ 70

        -2           1                                 3, 5

------------------------

                      3              5        0

                      3              5       0

                -   --------------------------------

                     0                0        0

6 0
2 years ago
Find the volume of the prism.
umka21 [38]

Answer:

C = 210cm^3

Step-by-step explanation:

We can get an estamate when we use the equation L x H x W = x/2

This shows 8 x 8 x 7,  then divide by 2 = 224cm^3

The reason it is slightly less could be the facts the teachers want you to remember this one.

For right side prism the equation is (1/2)b x h x  H

which means;

The volume of a triangular prism can be found by multiplying the base times the height. Both of the pictures of the Triangular prisms below illustrate the same formula. The formula, in general, is the area of the base (the red triangle in the picture on the right) times the height,h.

We use the side and base that surrounds the right side as (1/2)base and as side to find area then use the H as the height being the 7.

3 0
3 years ago
Read 2 more answers
Box of 15 gadgets is known to contain 5 defective gadgets if 4 gadgets are drawn at random what is the probability of finding no
baherus [9]

To solve this problem, we make use of the Binomial Probability equation which is mathematically expressed as:

P = [n! / r! (n – r)!] p^r * q^(n – r)

where,

n = the total number of gadgets = 4

r = number of samples = 1 and 2 (since not more than 2)

p = probability of success of getting a defective gadget

q = probability of failure = 1 – p

 

Calculating for p:

p = 5 / 15 = 0.33

So,

q = 1 – 0.33 = 0.67

 

Calculating for P when r = 1:

P (r = 1) = [4! / 1! 3!] 0.33^1 * 0.67^3

P (r = 1) = 0.3970

 

 

Calculating for P when r = 2:

P (r = 2) = [4! / 2! 2!] 0.33^2 * 0.67^2

P (r = 2) = 0.2933

 

Therefore the total probability of not getting more than 2 defective gadgets is:

P = 0.3970 + 0.2933

P = 0.6903

 

Hence there is a 0.6903 chance or 69.03% probability of not getting more than 2 defective gadgets.

5 0
3 years ago
Classify COB in the image above as either acute, obtuse, right, or straight angle.
Elis [28]

Answer:

Number 2 obtuse

Step-by-step explanation:

7 0
3 years ago
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In each part, decide if the statement is true. If it is true, prove it. If it is not true, give an explicit counterexample and d
svet-max [94.6K]

Answer: Hello!

ok, remember that "if and only if" implies that you need to prove the statement in both ways, this is represented with the ⇔ usually.

a) For any sets A, B and C, A ∩ B ⊆ C if and only if either A ⊆ C or B ⊆ C.

In this type of problems, i  find very useful start looking for some counterexample.

In this case, suppose that A = {1,2,3,4,5} , B = {3,4,5,6,7} and C = {3,4,5,6}

then is easy to see that A ⊄ C and B ⊄C.

And A∩B = {3,4,5}

then A∩B ⊂ C

then the statement is false (because one of the ways is false, remember that this is an "if and only if" statement)

b) For any sets A, B and C, A ⊆ B ∩ C if and only if both A ⊆ B and A ⊆ C.

the first way is true; because if A ⊆ B ∩ C. then all the elements of A are in the intersection of B and C (which are common elements for B and C) and then all the elements of A are in the set B and in the set C, and this means that A ⊆ B and A ⊆ C.

But let's see the other way now, suppose that A ⊆ B and A ⊆ C, now we want to know if A ⊆ B ∩ C.

if A ⊆ B and A ⊆ C, means that all the elements of A are in B, and all the elements of A are in C, then all the elements of A are common elements between B and C, this means that B ∩ C is at least equal to A (at least, because we know that all the elements of A are common elements between B and C, but there could be more common elements that don belong to A)

then A ⊆ B ∩ C.

4 0
3 years ago
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