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klemol [59]
3 years ago
6

What pattern is there in the power of 3

Mathematics
1 answer:
STALIN [3.7K]3 years ago
6 0
Guess its nine hundred and eighty seven


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If tñ=2ñ+3, find S10​
Allisa [31]

Answer:

S_{10}=280

Step-by-step explanation:

Given that,

t_n=2n+3 ...(1)

We need to find the value of S_{10}.

Put n = 1 to find the first term.

t_1=2(1)+3=5

Put n = 2 to find the second term.

t_2=2(2)+3=7

Put n = 3 to find the third term.

t_3=2(3)+3=9

Put n = 10 to find the tenth term.

t_{10}=2(10)+3=23

It means we need to find the sum of 5,7,9,.....,23.

The formula for the sum of n terms is given by :

S_=\dfrac{n}{2}(a+a_n)

We have, n = 10, a = 5 and a_n=23

So,

S=\dfrac{10}{2}(5+23)\\\\S_{10}=10\times 28\\\\=280

So, the value of S_{10} is equal to 280.

3 0
3 years ago
Which are vertical angles?<br> O AFE and BFD<br> O BFC and DFE<br> O AFE and CFD<br> O BFC and EFA
mel-nik [20]

the correct answer is a, because a vertical angle is made of 2 intercepting lines.

3 0
3 years ago
How to solve similar integer
SCORPION-xisa [38]
Ok, I'll make it simple for you: Keep Change Change. For all subtraction problems (and only subtraction, no addition), all you have to do is keep the first number, then change the sign from a minus to a plus, and then change the last number to the opposite. For example:
7 - (-3)
7 + 3=10

-4 - 10
-4 + -10= -14

For addition, if both numbers are negative, you just add them together. For example:
-21 + (-13)= -34

In this case you switch the sign and the number's sign also:
7 + (-18)
7 - 18 = -11

I hope I was help!!!
6 0
3 years ago
Suppose a random variable x is best described by a uniform probability distribution with range 22 to 55. Find the value of a tha
const2013 [10]

Answer:

(a) The value of <em>a</em> is 53.35.

(b) The value of <em>a</em> is 38.17.

(c) The value of <em>a</em> is 26.95.

(d) The value of <em>a</em> is 25.63.

(e) The value of <em>a</em> is 12.06.

Step-by-step explanation:

The probability density function of <em>X</em> is:

f_{X}(x)=\frac{1}{55-22}=\frac{1}{33}

Here, 22 < X < 55.

(a)

Compute the value of <em>a</em> as follows:

P(X\leq a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.95\times 33=[x]^{a}_{22}\\\\31.35=a-22\\\\a=31.35+22\\\\a=53.35

Thus, the value of <em>a</em> is 53.35.

(b)

Compute the value of <em>a</em> as follows:

P(X< a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.49\times 33=[x]^{a}_{22}\\\\16.17=a-22\\\\a=16.17+22\\\\a=38.17

Thus, the value of <em>a</em> is 38.17.

(c)

Compute the value of <em>a</em> as follows:

P(X\geq  a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.85=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.85\times 33=[x]^{55}_{a}\\\\28.05=55-a\\\\a=55-28.05\\\\a=26.95

Thus, the value of <em>a</em> is 26.95.

(d)

Compute the value of <em>a</em> as follows:

P(X\geq  a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.89=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.89\times 33=[x]^{55}_{a}\\\\29.37=55-a\\\\a=55-29.37\\\\a=25.63

Thus, the value of <em>a</em> is 25.63.

(e)

Compute the value of <em>a</em> as follows:

P(1.83\leq X\leq  a)=\int\limits^{a}_{1.83} {\frac{1}{33}} \, dx \\\\0.31=\frac{1}{33}\cdot \int\limits^{a}_{1.83} {1} \, dx \\\\0.31\times 33=[x]^{a}_{1.83}\\\\10.23=a-1.83\\\\a=10.23+1.83\\\\a=12.06

Thus, the value of <em>a</em> is 12.06.

7 0
3 years ago
Are these triangles similar?
Sliva [168]
Yes they are similar

One triangle is just smaller but they are both triangles
7 0
3 years ago
Read 2 more answers
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