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adoni [48]
4 years ago
14

Explain how a calculator shows the remainder when you divide 14 5 by 8

Mathematics
2 answers:
Vladimir [108]4 years ago
8 0
If you put 145 divided by 8 the remainder on a calculator would come out as a decimal
hjlf4 years ago
3 0
It would equal 18.125
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How do I solve this?
Ivenika [448]

\overline{XS}\cong\overline{YT}\Rightarrow 3m+7=4.2m+5\ \ \ |-7\\\\3m=4.2m-2\ \ \ \ |-4.2m\\\\-1.2m=-2\ \ \ \ |:(-1.2)\\\\m=\dfrac{2}{1.2}\\\\m=\dfrac{20}{12}\\\\m=\dfrac{5}{3}\to m=1\dfrac{2}{3}


\overline{YS}\cong\overline{XT}\Rightarrow3\dfrac{1}{2}r+2=2r+5\ \ \ \ |-2\\\\3\dfrac{1}{2}r=2r+3\ \ \ \ |-2r\\\\1\dfrac{1}{2}r=3\ \ \ |\cdot2\\\\3r=6\ \ \ \ |:3\\\\r=2

7 0
3 years ago
Tom plants 3 seeds.
Rasek [7]

Answer:

A. 64/125

B. 124/125.

Step-by-step explanation:

A).  As the events ( germinate or not germinate) are independent we multiply the probabilities.

Prob(All seeds germinate) = 4/5*4/5*4/5 =  64/125.

B). Probability of at least one germinating =  1 - probability that none germinate

Probability of  1 seed not germinating = 1 -45 = 1/5.

So Prob(at least one germinating)

= 1 - (1/5 * 1/5 * 1/5)

= 1 - 1/125

= 124/125.

5 0
3 years ago
Which names are correct for FM−→− ?
fiasKO [112]

Answer with explanation:

⇒A segment , \Bar {AB} is equivalent to \Bar {BA}.

⇒We can assert this by this way, if Distance between Two points A and B is 6 unit, then Distance between B and A will be 6 unit also.

So,

\bar{FM}=\bar{MF}

FM−→−⇒MF−→−

Option A

6 0
4 years ago
Read 2 more answers
Find the mode of the following data:
frutty [35]

Answer:

Step-by-step explanation:

The data which has higher frequency in the given data set is the mode.

Mode = 33

5 0
2 years ago
Suppose that we have a sample space S5 {E1, E2, E3, E4E4E, E5, E6, E7}, where E1, E2, . . ., E7 denote the sample points. The fo
VashaNatasha [74]

Answer:

P(A) = 0.4 ; P(B) = 0.50 ; P(C) = 0.60 ; P(A u B) = 0.65 ; P(A n B) = 0.25 ;

A and C are mutually exclusive ;

0.5

Step-by-step explanation:

S = {E1, E2, E3, E4, E5, E6, E7}

P(E1) = .05, P(E2) = .20, P(E3) = .20, P(E4) = .25, P(E5) = .15, P(E6) = .10, and P(E7) = .05

Let:

A{E1,E4,E6}

B{E2,E4,E7}

C{E2,E3,E5,E7}

P(A) = 0.05 + 0.25 + 0.10 = 0.40

P(B) = 0.20 + 0.25 + 0.05 = 0.50

P(C) = 0.20 + 0.20 + 0.15 + 0.05 = 0.60

AUB = {E1, E2, E4, E6, E7}

P(A u B) = 0.05 + 0.20 + 0.25 + 0.10 + 0.05 = 0.65

AnB = {E4}

P(A n B) = 0.25

A and C are mutually exclusive if AnC = ∅

A n C = ∅

Hence, A and C are mutually exclusive.

B complement = B' = {E1, E3, E5, E6}

P(B') = 0.05 + 0.20 + 0.15 + 0.10 = 0.5

5 0
3 years ago
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