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givi [52]
3 years ago
13

BEST ANSWER GETS BRANLIEST AND 100 POINTS Simplify (√3+√48)(√2−√5).

Mathematics
2 answers:
Nadya [2.5K]3 years ago
7 0

Answer:

5\sqrt{6}-5\sqrt{15}

Step-by-step explanation:

(\sqrt{3} )(\sqrt{2} )+(\sqrt{48} )(\sqrt{2} )-(\sqrt{3} )(\sqrt{5} )-(\sqrt{48} )(\sqrt{5} )

\sqrt{6}+\sqrt{2*3*2*2*2*2}  -\sqrt{15}-\sqrt{2*3*2*2*2*5}

\sqrt{6}+4\sqrt{3*2} -\sqrt{15} -4\sqrt{3*5}

\sqrt{6}+4\sqrt{6} -\sqrt{15} -4\sqrt{15}

5\sqrt{6} -5\sqrt{15}

UNO [17]3 years ago
7 0

OOOOIIIIIII its 5/6- 5/6

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The town of Hayward (CA) has about 50,000 registered voters. A political research firm takes a simple random sample of 500 of th
timama [110]

Answer: (0.076, 0.140)

Step-by-step explanation:

Confidence interval for population proportion (p) is given by :-

\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

, where \hat{p} = sample proportion.

n= sample size.

\alpha = significance level .

z_{\alpha/2} = critical z-value (Two tailed)

As per given , we have

sample size : n= 500

The number of Independents.:  x= 54

Sample proportion of Independents\hat{p}=\dfrac{x}{n}=\dfrac{54}{500}=0.108

Significance level 98% confidence level : \alpha=1-0.98=0.02

By using z-table , Critical value : z_{\alpha/2}=z_{0.01}=2.33

The 98% confidence interval for the true percentage of Independents among Haywards 50,000 registered voters will be :-

0.108\pm (2.33)\sqrt{\dfrac{0.108(1-0.108)}{500}}\\\\=0.108\pm2.33\times0.013880634\\\\=0.108\pm0.03234187722\\\\\approx0.108\pm0.032=(0.108-0.032,\ 0.108+0.032)=(0.076,\ 0.140)

Hence, the 98% confidence interval for the true percentage of Independents among Haywards 50,000 registered voters.= (0.076, 0.140)

8 0
3 years ago
What is the distance between the points (7,-6) and (13,2)? Please show all work.
lana [24]

Answer:

Step-by-step explanation:

(7,-6) (13,2)

2--6. 8. 4

------= ------= ------ (2--6 turns to a +)

13-7. 6. 3

4

Slope:------

3

Points=(7,-6)

4

y--6=----(x-7) (y--6 turns to a +)

3

4. 28

y+6= ---- x - ----

3. 3

-6. -6

-------------------------

4. 46

y=----- x - -------

3. 3

6 0
3 years ago
32) (a)-(c) pre calculus
Harrizon [31]
Well, remember we can't take the square root of a negative

so we see that we have \sqrt{1-x^2}
so find those values that take sqrt of a negative and restrict hem from the domain
anny value greater than 1 and less than -1
so domain is from -1 to 1, including those numbers
D=[-1,1]

a. D=[-1,1] or from -1 to 1 is domain


b. for a TI-84, go to y-editor then input 4x \sqrt{1-x^2} for y1

c. for a TI-84, click 2nd then window (gets to tbset) scrol down to set Δx to 0.1, then cilick 2nd again then click graph (to select table) and scroll down till you see that value of y that is the biggest, that value is x=0.7



A. domain is from -1 to 1
B. use your brain or google the instructions for your calulator
C. at x=0.7


8 0
3 years ago
X - 10 > 3
joja [24]

Answer:

x>13

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has i
goldfiish [28.3K]

Answer:

  • P(t) = 100·2.3^t
  • 529 after 2 hours
  • 441 per hour, rate of growth at 2 hours
  • 5.5 hours to reach 10,000

Step-by-step explanation:

It often works well to write an exponential expression as ...

   value = (initial value)×(growth factor)^(t/(growth period))

(a) Here, the growth factor for the bacteria is given as 230/100 = 2.3 in a period of 1 hour. The initial number is 100, so we can write the pupulation function as ...

  P(t) = 100·2.3^t

__

(b) P(2) = 100·2.3^2 = 529 . . . number after 2 hours

__

(c) P'(t) = ln(2.3)P(t) ≈ 83.2909·2.3^t

  P'(2) = 83.2909·2.3^2 ≈ 441 . . . bacteria per hour

__

(d) We want to find t such that ...

  P(t) = 10000

  100·2.3^t = 10000 . . . substitute for P(t)

  2.3^t = 100 . . . . . . . . divide by 100

  t·log(2.3) = log(100)

  t = 2/log(2.3) ≈ 5.5 . . . hours until the population reaches 10,000

6 0
3 years ago
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