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3 years ago

6

Help I’m struggling so much in math

1 answer:

andreev551 [17]3 years ago

7 0

Slope = rise/run
rise/run = -8/5

FINAL ANSWER -8/5

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According to the graph above,what is the unit rate of a human's heartless per minute

kari74 [83]

The unit rate for the graph above is 60 heartbeats per minute. Half of 2 is 1 so go to the 1 min. mark and go up until you reach the line graph and it says 60. Do the same for the 2 min. mark and go up until you reach the line graph and it'll say 120. 60 ÷60=1 and 120 ÷ 60= 2

5 0

3 years ago

A fair coin is flipped twelve times. What is the probability of the coin landing tails up exactly nine times?

seraphim [82]

Answer:

$P\left(E\right)=\frac{55}{1024}$

Step-by-step explanation:

Given that a fair coin is flipped twelve times.

It means the number of possible sequences of heads and tails would be:

2¹² = 4096

We can determine the number of ways that such a sequence could contain exactly 9 tails is the number of ways of choosing 9 out of 12, using the formula

$nCr=\frac{n!}{r!\left(n-r\right)!}$

Plug in n = 12 and r = 9

$=\frac{12!}{9!\left(12-9\right)!}$

$=\frac{12!}{9!\cdot \:3!}$

$=\frac{12\cdot \:11\cdot \:10}{3!}$ ∵ $\frac{12!}{9!}=12\cdot \:11\cdot \:10$

$=\frac{1320}{6}$ ∵ $3!\:=\:3\times 2\times 1=6$

$=220$

Thus, the probability will be:

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$

$=\frac{220}{4096}$

$=\frac{55}{1024}$

Thus, the probability of the coin landing tails up exactly nine times will be:

$P\left(E\right)=\frac{55}{1024}$

4 0

3 years ago

What does (26/13).(-7)+14-4=

Bond [772]

Answer:

-4

Step-by-step explanation:

1. Always start with the parenthesis. (26/13) = 2

2. Then multiply. 2*-7 = -14

3. After, you add 14 to your precious answer (-14) then subtract 4.

-14 + 14 = 0 - 4 = -4

5 0

3 years ago

Question points)

mash [69]

Answer:

listen to lil peep ghost girl and I'll answer this for you

Step-by-step explanation:

just playing with u i tried but i can't find the darn answer

5 0

3 years ago

Find the smallest positive $n$ such that \begin{align*} n &\equiv 3 \pmod{4}, \\ n &\equiv 2 \pmod{5}, \\ n &\equiv

Alex777 [14]

4, 5, and 7 are mutually coprime, so you can use the Chinese remainder theorem right away.

We construct a number $x$ such that taking it mod 4, 5, and 7 leaves the desired remainders:

$x=3\cdot5\cdot7+4\cdot2\cdot7+4\cdot5\cdot6$

Taken mod 4, the last two terms vanish and we have

$x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod4$

so we multiply the first term by 3.

Taken mod 5, the first and last terms vanish and we have

$x\equiv4\cdot2\cdot7\equiv51\equiv1\pmod5$

so we multiply the second term by 2.

Taken mod 7, the first two terms vanish and we have

$x\equiv4\cdot5\cdot6\equiv120\equiv1\pmod7$

so we multiply the last term by 7.

Now,

$x=3^2\cdot5\cdot7+4\cdot2^2\cdot7+4\cdot5\cdot6^2=1147$

By the CRT, the system of congruences has a general solution

$n\equiv1147\pmod{4\cdot5\cdot7}\implies\boxed{n\equiv27\pmod{140}}$

or all integers $27+140k$ , $k\in\mathbb Z$ , the least (and positive) of which is 27.

3 0

3 years ago