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Darya [45]
3 years ago
9

A spinner of 10 divisions, numbered from 1 to 10, is spun and two coins are tossed. What is the probability to land on a prime n

umber in the spin and getting two tails from the two tosses?
Mathematics
1 answer:
pogonyaev3 years ago
6 0

Answer:

1/10

Step-by-step explanation:

There are 4 prime numbers from 1 to 10 (2,3,5,7). This makes the probability of getting a prime number as 4/10 = 2/5.

The probability of getting one tail is 1/2, making the proability of getting 2 to be (1/2)^2 = 1/4.

Multiply both probabilities together, and you get 2/5 * 1/4 = 1/10

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Use properties of exponents to complete the equivalent expression. (2^3)^3 = (2 · 2 · 2)^3 = (2 · 2 · 2) · ( ? · ? · ? ) (2 · 2
garri49 [273]
Thats confusing??? holy cow
5 0
3 years ago
Read 2 more answers
PLS HELP ASAP! WILL MARK BRAINLIEST<br> sin ∠A =<br><br> 3/5<br> 4/5<br> 3/4<br> 5/4
Dafna11 [192]

Answer:

3/5

Step-by-step explanation:

sohcahtoa

since we're finding sine, we need the opposite and hypotenuse.

opposite of <A=3, hyp of <A=5.

opp comes before hyp in "soh", so sin<A=3/5

5 0
3 years ago
Find the measure of a positive angle coterminal with - 240°.
Irina18 [472]
-240+360=120
The answer is 120 degrees.
5 0
3 years ago
SOLUTION We observe that f '(x) = -1 / (1 + x2) and find the required series by integrating the power series for -1 / (1 + x2).
Ann [662]

Answer:

Required series is:

\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....

Step-by-step explanation:

Given that

                           f'(x) = -\frac{1}{1 + x^{2}} ---(1)

We know that:

                  \frac{d}{dx}(tan^{-1}x)=\frac{1}{1+x^{2}} ---(2)

Comparing (1) and (2)

                           f'(x)=-(tan^{-1}x) ---- (3)

Using power series expansion for tan^{-1}x

f'(x)=-tan^{-1}x=-\int {\frac{1}{1+x^{2}} \, dx

= -\int{ \sum\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx

= -\sum{ \int\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx

=-[c+\sum\limits^{ \infty}_{n=0} (-1)^{n}\frac{x^{2n+1}}{2n+1}]

=C+\sum\limits^{ \infty}_{n=0} (-1)^{n+1}\frac{x^{2n+1}}{2n+1}

=C-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....

as

                 tan^{-1}(0)=0 \implies C=0

Hence,

\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....

7 0
3 years ago
What’s the correct answer for this question?
Zolol [24]

Answer:

Absolute change: $1,700

Relative change: 7.6577%

Step-by-step explanation:

Absolute change: $22,200 - $20,500 = $1,700

Relative change: (100 × $1,700) : $22,200 ≈ 7.6577%

6 0
2 years ago
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