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BartSMP [9]
4 years ago
15

iddle" class="latex-formula">
Mathematics
1 answer:
const2013 [10]4 years ago
4 0

8x^2=2x+1\ \ \ \ |-2x-1\\\\8x^2-2x-1=0\\\\8x^2+2x-4x-1=0\\\\2x(4x+1)-1(4x+1)=0\\\\(4x+1)(2x-1)=0\iff4x+1=0\ \vee\ 2x-1=0\\\\4x+1=0\ \ \ |-1\\4x=-1\ \ \ |:4\\x=-0.25\\\\2x-1=0\ \ \ |+1\\2x=1\ \ \ \ |:2\\x=0.5\\\\Answer:\ x=-0.25\ or\ x=0.5

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The sum of three numbers is 18. If every number is a prime number, what are the three numbers?
sleet_krkn [62]
The three numbers are 2, 5, and 11.

The sum of 3 numbers is 18.        2 + 5 + 11 = 18

Since the problem states that every number is a prime number, then the number must be a natural number greater than 1 that has no positive divisors other than 1 and itself.

2 / 1 = 2  OR  2 / 2 = 1  These numbers are prime numbers because their divisors
5 / 1 = 5  OR  5 / 5 = 1            are 1 and itself.
11/1 = 11 OR 11/11 = 1
4 0
3 years ago
Read 2 more answers
Math home work how do I find the area I also need the work please help meeeeeeee
AVprozaik [17]
For the shaded bit all you need to do is 8.5 × 6.
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4 0
4 years ago
What is the area of the figure?
luda_lava [24]
The area is 12 feet
4 0
3 years ago
Alonso went to the market with \$55$55dollar sign, 55 to buy eggs and sugar. He knows he needs a package of 121212 eggs that cos
Neko [114]

Answer: See explanation

Step-by-step explanation:

Based on the scenario in the question, the expression to calculate the number of boxes of sugar Alonso can buys will be:

= 2.75 + 11.50S ≤ 55

When solved further, this will be:

2.75 + 11.50S ≤ 55

11.50S ≤ 55 - 2.75

11.50S ≤ 52.25

S ≤ 52.25 / 11.50

S ≤ 4.54

He can buy 4 boxes of sugar

3 0
3 years ago
Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.7
Zinaida [17]

Answer:

(a) The probability that the sample average sediment density is at most 3.00 is 0.092.

    The probability that the sample average sediment density is between 2.70 and 3.00 is 0.477.

(b) The sample size must be at least 77.

Step-by-step explanation:

The random variable <em>X</em> ca be defined as the sediment density (g/cm) of a specimen from a certain region.

The random variable <em>X</em> is normally distributed with mean, <em>μ</em> = 2.7 and standard deviation, <em>σ</em> = 0.75.

(a)

A random sample of <em>n</em> = 25 specimens is selected.

Compute the probability that the sample average sediment density is at most 3.00 as follows:

Apply continuity correction:

P(\bar X\leq 3.00)=P(\bar X

                    =P(\bar X

                    =P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}

                    =P(Z

Thus, the probability that the sample average sediment density is at most 3.00 is 0.092.

Compute the probability that the sample average sediment density is between 2.70 and 3.00 as follows:

P(2.70

                                =P(0

*Use a <em>z</em>-table.

Thus, the probability that the sample average sediment density is between 2.70 and 3.00 is 0.477.

(b)

It is provided that:

P(\bar X\leq 3.00)\geq 0.99

P(\bar X

P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}

P(Z

The value of <em>z</em> for the probability above is

<em>z</em> ≥ 2.33

Compute the value of <em>n</em> as follows:

\frac{|2.50-2.70|}{0.75/\sqrt{n}}\geq 2.33

\frac{|-0.20|}{2.33}\geq \frac{0.75}{\sqrt{n}}

\sqrt{n}\geq \frac{0.75\times 2.33}{|-0.20|}\\\sqrt{n}\geq 8.7375\\n\geq 76.3439\\\approx n\geq 77

Thus, the sample size must be at least 77.

8 0
3 years ago
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