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stira [4]
3 years ago
15

40 POINTS!!!! What is the value x? 15 25 30 35

Mathematics
1 answer:
harina [27]3 years ago
7 0

Answer:

35

Step-by-step explanation:

If the two lines are parallel and has other line intersect ; the angle will be equal so the answer is 35.

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Ax + b2 = cx + d2<br> ????
ryzh [129]

Answer:

x=b^2+d^2/a-c

Step-by-step explanation:

5 0
3 years ago
Find the sum of a finite geometric sequence from n = 1 to n = 7, using the expression −4(6)n − 1.
Verizon [17]

Answer:

<h2>-223,948</h2>

Step-by-step explanation:

The formula of a sum of terms of a gometric sequence:

S_n=a_1\cdot\dfrac{1-r^n}{1-r}

a₁ - first term

r - common ratio

We have

a_n=-4(6)^{n-1}

Calculate a₁. Put n = 1:

a_1=-4(6)^{1-1}=-4(6)^0=-4(1)=-4

Calculate the common ratio:

r=\dfrac{a_{n+1}}{a_n}\\\\a_{n+1}=-4(6)^{n+1-1}=-4(6)^n\\\\r=\dfrac{-4(6)^n}{-4(6)^{n-1}}=6^n:6^{n-1}\\\\\text{use}\ a^n:a^m=a^{n-m}\\\\r=6^{n-(n-1)}=6^{n-n+1}=6^1=6

\text{Substitute}\ a_1=-4,\ n=7,\ r=6:\\\\S_7=-4\cdot\dfrac{1-6^7}{1-6}=-4\cdot\dfrac{1-279936}{-5}=-4\cdot\dfrac{-279935}{-5}=(-4)(55987)\\\\S_7=-223948

7 0
3 years ago
Suppose the number of insect fragments in a chocolate bar follows a Poisson process with the expected number of fragments in a 2
leonid [27]

Answer:

a)The expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b)0.6004

c)19.607

Step-by-step explanation:

Let X denotes the number of fragments in 200 gm chocolate bar with expected number of fragments 10.2

X ~ Poisson(A) where \lambda = \frac{10.2}{200} = 0.051

a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar

\frac{1}{4} \times 200 = 50

50 grams of bar contains expected fragments = \lambda x = 0.051 \times 50=2.55

So, the expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b) Now we are supposed to find the probability that you have to eat more than 10 grams of chocolate bar before ending your first fragment

Let X denotes the number of grams to be eaten before another fragment is detected.

P(X>10)= e^{-\lambda \times x}= e^{-0.051 \times 10}= e^{-0.51}=0.6004

c)The expected number of grams to be eaten before encountering the first fragments :

E(X)=\frac{1}{\lambda}=\frac{1}{0.051}=19.607 grams

7 0
3 years ago
The bird population on an island is declining at a rate of 2.2% per year. The population was 3500 in the year 2016.
Elza [17]
The correct answer is C
8 0
3 years ago
Read 2 more answers
2:1<br> and <br> 18:9<br> equivalent?
denis23 [38]
Yes they are equivalent 18:9 simplified by 9 is 2:1
8 0
3 years ago
Read 2 more answers
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