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3 years ago

X + 26 = 39; x = I NEED HELP !!

Mathematics

2 answers:

Serjik [45]3 years ago

8 0

Answer:

x=13

i accadently put 36, but you can still see how it happend

Step-by-step explanation:

Think bout brainliest thank you

pashok25 [27]3 years ago

3 0

13

step by step explanation: subtract 26 on each side and u get 13

hope this helps

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PLEASE HELP ASAP TIMED

Kay [80]

Answer:

D ? I think

Step-by-step explanation:

7 0

3 years ago

Suppose that start-up companies in the area of biotechnology have probability 0.2 of becoming profitable, and that those in the

schepotkina [342]

Answer:

a) 0.03

b) 0.68

c) 0.32

Step-by-step explanation:

We are given the following in the question:

B: companies in the area of biotechnology

I: companies in the area of information technology

P(B) = 0.2

P(I) = 0.15

The two events are given to be independent.

a) P(both companies become profitable)

$P(B\cap I) = P(B)\times P(I)\\= 0.2\times 0.15 = 0.03$

0.03 is the probability that both companies become profitable

b) P(neither company becomes profitable)

$P(B'\cap I')\\=P(B')\times P(I')\\=(1-P(B))(1-P(I))\\=(1-0.2)(1-0.15)\\=0.68$

0.68 is the probability that neither company becomes profitable.

c) P(at least one of the two companies become profitable)

$P(B\cup I) = P(B) + P(I) - P(B\cap I)\\= 0.2 + 0.15 - 0.03\\=0.32$

0.32 is the probability that at least one of the two companies become profitable

4 0

3 years ago

Given $a \equiv 1 \pmod{7}$, $b \equiv 2 \pmod{7}$, and $c \equiv 6 \pmod{7}$, what is the remainder when $a^{81} b^{91} c^{27}$

Blizzard [7]

$\begin{cases}a\equiv1\pmod7\\b\equiv2\pmod7\\c\equiv6\pmod7\end{cases}$

$a^{81}\equiv1^{81}\equiv1\pmod7$

$b^{91}\equiv2^{91}\pmod7$

$2^{91}\equiv(2^3)^{30}\times2^1\equiv8^{30}\times2\pmod7$
$8\equiv1\pmod7$
$2\equiv2\pmod7$
$\implies2^{91}\equiv1^{30}\times2\equiv2\pmod7$

$c^{27}\equiv6^{27}\pmod7$

$6^{27}\equiv(-1)^{27}\equiv-1\equiv6\pmod7$

$\implies a^{81}b^{91}c^{27}\equiv1\times2\times6\equiv12\equiv5\pmod7$