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Wewaii [24]
3 years ago
5

Can you please help me with my question​

Mathematics
1 answer:
Reika [66]3 years ago
4 0
The answer:
First: sec(-pi) is sec(pi)
And sec(pi) is -1
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Brian’s school locker has a three-digit combination lock that can be set using the numbers 5 to 9 (including 5 and 9), without r
andreev551 [17]
The possible digits are: 5, 6, 7, 8 and 9. Let's mark the case when the locker code begins with a prime number as A and the case when <span>the locker code is an odd number as B. We have 5 different digits in total, 2 of which are prime (5 and 7).

First propability:
</span>P_A=\frac{2}{5}=40\%
<span>
By knowing that digits don't repeat we can say that code is an odd number in case it ends with 5, 7 or 9 (three of five digits).

Second probability:
</span>P_B=\frac{3}{5}=60\%
5 0
3 years ago
If 2k, 5k-1 and 6k+2 are the first 3 terms of an arithmetic sequence, find k and the 8th term​
Klio2033 [76]

Answer:

see explanation

Step-by-step explanation:

The common difference d of an arithmetic sequence is

d = a_{2} - a_{1} = a_{3} - a_{2}

Substitute in values and solve for k, that is

5k - 1 - 2k = 6k + 2 - (5k - 1)

3k - 1 = 6k + 2 - 5k + 1

3k - 1 = k + 3 ( subtract k from both sides )

2k - 1 = 3 ( add 1 to both sides )

2k = 4 ⇒ k = 2

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

The n th term of an arithmetic sequence is

a_{n} = a_{1} + (n - 1)d

a_{1} = 2k = 2 × 2 = 4 and

d = 5k - 1 - 2k = 3k - 1 = (3 × 2) - 1 = 5

Hence

a_{8} = 4 + (7 × 5) = 4 + 35 = 39

4 0
3 years ago
PLEASE HELP ME!!!!!!
Sonja [21]

Answer:

\huge \red{ c_3 = - 1}

Step-by-step explanation:

c_1 = 1 \\  \\ c_n= - 2c_{n-1}+5 \\  \\ c_2 = - 2c_{2-1}+5 \\  \\ c_2 = - 2c_{1}+5 \\  \\ c_2 = - 2(1)+5 \\  \\ c_2 = - 2+5 \\  \\  \huge \purple{c_2 =3} \\  \\ c_3 = - 2c_{3-1}+5 \\  \\ c_3 = - 2c_{2}+5 \\  \\ c_3 = - 2(3)+5 \\  \\ c_3= - 6+5 \\  \\ \huge \red{ c_3 = - 1}

6 0
3 years ago
measurement error that is continuous and uniformly distributed from millivolts is added to the true voltage of a circuit. then t
satela [25.4K]

Question has missing details (Full question below)

Measurement error that is continuous and uniformly distributed from –3 to +3 millivolts is added to a circuit’s true voltage. Then the measurement is rounded to the nearest millivolt so that it becomes discrete. Suppose that the true voltage is 219 millivolts. What is the mean and variance of the measured voltage

Answer:

Mean = 219

Variance = 4

Step-by-step explanation:

Given

Let X be a random variable measurement error.

X has a discrete uniform distribution as follows

a = 219 - 3 = 216

b = 219 + 3 = 222

Mean or Expected value is calculated as follows;

E(x) = ½(216+222)

E(x) = ½ * 438

E(x) = 219

Variance is calculated as follows;

Var(x) = ((b-a+1)²-1)/12

Var(x) = ((222-216+1)²-1)/12

Var(x) = (7²-1)/12

Var(x) = 48/12

Var(x) = 4

8 0
3 years ago
If you had 12 pieces of licorice to share equally among 5 people, how much licorice would each
pentagon [3]
12÷5=1.5 of licorice will each person get
7 0
3 years ago
Read 2 more answers
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