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mojhsa [17]
3 years ago
11

Can someone help me with 39 and 40?

Mathematics
2 answers:
loris [4]3 years ago
6 0
39.
9.2 x 10^3, 10,000, 1.8 x 10^4


40.
0.005, 5.25 x10^-3, 5.0 x 10^-2
faltersainse [42]3 years ago
6 0

39. Since exponents with 10 give us the number of zeros we are working with ex.(10^3= 1000, 3 zeros) We can just find out what we need to multiply and we can order them. 9.2x 1000= 9200. Now we just need to find the last one, being 10^4= 10,000x 1.8= 18,000. Now we order them: 9,200, 10,000, 18,000.


40. The same rule about working with exponents can be said when working with negative exponents but instead of how many zeros we add, it's how many zeros we take a step to the left from (ex. 10^-4= 0.0001, 4 zeros back.) So we just find the exponents' real value and work with them. 5.25 x 0.001= 0.00525. 5 x 0.01= 0.05. Now we order them: 0.005, 0.00525, 0.05

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Answer:

(1) The possible outcomes are: X = {0, 1, 2, 3}.

(2) The number of times should Hartley spin a difference of 1 is 36.

(3) The number of times should Hartley spin a difference of 0 is 24.

Step-by-step explanation:

The number of sections on the spinner is 4 labelled as {1, 2, 3, 4}.

The total number of spins for each of the spinner is, <em>n</em> = 96.

(1)

The sample space of spinning both the spinners together are:

S = {(1, 1), (1, 2), (1, 3), (1, 4)

      (2, 1), (2, 2), (2, 3), (2, 4)

      (3, 1), (3, 2), (3, 3), (3, 4)

      (4, 1), (4, 2), (4, 3), (4, 4)}

Total = 16.

The possible outcomes are:

X = {0, 1, 2, 3}.

(2)

The sample space with the difference 1 are:

S₁ = {(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3)}

n (S₁) = 6

The probability of the difference 1 is:

P(\text{Diff}=1)=\frac{n(S_{1})}{N}=\frac{6}{16}=\frac{3}{8}

The spinners were spinner 96 times.

The expected number of times would Hartley spin a difference of 1 is:

E(\text{Diff}=1)=P(\text{Diff}=1)\times n\\\\=\frac{3}{8}\times 96\\\\=36

Thus, the number of times should Hartley spin a difference of 1 is 36.

(3)

The sample space with the difference 0 are:

S₂ = {(1, 1), (2, 2), (3, 3), (4, 4)}

n (S₂) = 4

The probability of the difference 0 is:

P(\text{Diff}=0)=\frac{n(S_{2})}{N}=\frac{4}{16}=\frac{1}{4}

The spinners were spinner 96 times.

The expected number of times would Hartley spin a difference of 0 is:

E(\text{Diff}=0)=P(\text{Diff}=0)\times n\\\\=\frac{1}{4}\times 96\\\\=24

Thus, the number of times should Hartley spin a difference of 0 is 24.

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