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3 years ago

15

Given that

="latex-formula"> is it ever possible to have $\sqrt{a} +\sqrt{b} =\sqrt{a+b}$ ? Explain.

1 answer:

xz_007 [3.2K]3 years ago

8 0

Answer:

Step-by-step explanation:

√a+√b=√(a+b)

squaring

a+b+2√a√b=a+b

or 2√a√b=0

which is possible only if a=0 or b=0

Both a and b can't be zero as a≠b

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The total cost C of bulk jelly beans varies directly as the number of pounds p. If

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Answer:

it would be 4 pounds

Step-by-step explanation:

8.98 times 4 equals 35.92

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2 years ago

A triangle has a 90 degree angle and two sides that measure 5cm in length. Read the statements

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Yes (True)

Step-by-step explanation:

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3 years ago

Given an integer, how do you find it's opposite?

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3 years ago

A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as

Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

Step-by-step explanation:

Let <em>X</em> = number of seconds in the batch.

The probability of the random variable <em>X</em> is, <em>p</em> = 0.31.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

$P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...$

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

$P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888$

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

$=1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776$

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

$={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453$

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

8 0

3 years ago

The senior class want to play a practical joke on the principal, who was a former college basketball star. They want to fill his

steposvetlana [31]

The number of basketball that will fill up the entire office is <u>approximately 16,615.</u>

<em><u>Recall:</u></em>

Volume of a spherical shape = $\frac{4}{3} \pi r^3$

Volume of a rectangular prism = $l \times w \times h$

<em><u>Given:</u></em>

Diameter of basketball = 9.5 in.

Radius of the ball = 1/2 of 9.5 = 4.75 in.

Radius of the ball in ft = 0.4 ft (12 inches = 1 ft)

Dimension of the office (rectangular prism) = 20 ft by 18 ft by 12 ft

First, find the volume of the basketball:

Volume of ball = $\frac{4}{3} \pi r^3 = \frac{4}{3} \times \pi \times 4.75^3\\$

Volume of basketball = $448.92 $ in^3$

Convert to $ft^3$

$1728 $ in.^3 = 1 $ ft^3$

<em>Therefore,</em>

Volume of basketball = $\frac{448.92}{1728} = 0.26 $ ft^3$

Find the volume of the office (rectangular prism):

Volume of the office = $20 \times 18 \times 12 = 4,320 $ ft^3$

Number of basket ball that will fill the office = Volume of office / volume of basketball

<em>Thus:</em>

Number of basket ball that will fill the office = $\frac{4,320}{0.26} = 16,615$

Therefore, it will take approximately <u>16,615 balls</u><u> to fill up the entire office</u>.

Learn more here:

brainly.com/question/16098833

6 0

2 years ago