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3 years ago

3) х 45° 11 A) 10.8 C) 11.0 B) 11.6 D) 8.8

Mathematics

1 answer:

maxonik [38]3 years ago

6 0

Answer:

AAO 22.902 A 45 12.632 13,939 945 559,302 5A , 480 1.17 6.044 36,491 ... 12.1 F 13.2 12.1 F 13.2 12.1 E 13.2 12.1 F 12.9 12.5 8.6 8.3 F 10.0 9.7 8.6 B.3 E 11.6 ... 10.5 D 11.8 10.5 C 11.8 10.5 D 11.8 10.5 с 11.8 10.5 U 11.8 10.5 D 13.3 12.7 ... 10.1 12.1 11.4 8.8 A.3 13.3 12.9 13.3 12.5 13.3 12.5 13.3 12.5 11.0 10.4 13.3 ...

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Find a round to the nearest tenth a 150 12 10cm c a=?cm

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Answer:

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3 years ago

Use the quadratic formula to find the solutions to the quadratic equation below 2x^2-5x+5=0

Serggg [28]

Answer:

x = (5 + i sqrt(15))/4 or x = (5 - i sqrt(15))/4

Step-by-step explanation:

Solve for x:

2 x^2 - 5 x + 5 = 0

Hint: | Using the quadratic formula, solve for x.

x = (5 ± sqrt((-5)^2 - 4×2×5))/(2×2) = (5 ± sqrt(25 - 40))/4 = (5 ± sqrt(-15))/4:

x = (5 + sqrt(-15))/4 or x = (5 - sqrt(-15))/4

Hint: | Express sqrt(-15) in terms of i.

sqrt(-15) = sqrt(-1) sqrt(15) = i sqrt(15):

Answer: x = (5 + i sqrt(15))/4 or x = (5 - i sqrt(15))/4

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3 years ago

9<br> Which expression is equivalent to 181 + |-7|?<br> -15<br> -1<br> 1<br> 15

pochemuha

Answer:

Step-by-step explanation:

|-7|=7

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3 years ago

An elementary school class ran 1 mile in an average of 11 minutes with a standard deviation of 3 minutes. Rachel, a student in t

EleoNora [17]

Answer:

Rachel

Step-by-step explanation:

We need to measure how far (towards the left) are the students from the mean in<em> “standard deviations units”</em>.

That is to say, if t is the time the student ran the mile and s is the standard deviation of the class, we must find an x such that

mean - x*s = t

For Rachel we have

11 - x*3 = 8, so x = 1.

Rachel is <em>1 standard deviation far (to the left) from the mean</em> of her class

For Kenji we have

9 - x*2 = 8.5, so x = 0.25

Kenji is <em>0.25 standard deviations far (to the left) from the mean</em> of his class

For Nedda we have

7 - x*4 = 8, so x = 0.25

Nedda is also 0.25 standard deviations far (to the left) from the mean of his class.

As Rachel is the farthest from the mean of her class in term of standard deviations, Rachel is the fastest runner with respect to her class.

8 0

3 years ago

Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand

dybincka [34]

Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

$P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...$

Thus $P(E)$ = $\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!$

Thus our probability becomes

$P(E)=\frac{1}{\binom{r+w}{r}}$

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at $(r+1)^{th}$ draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals $\binom{r}{r-1}$

Thus the probability becomes $P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}$

Thus required probability of case b becomes $P(E)+ P(E_{i})$

= $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\$

7 0

3 years ago