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3 years ago

The area of a rectangle is 27 cm. Its length is 9 cm. What is the width of the rectangle?

Mathematics

2 answers:

butalik [34]3 years ago

5 0

Answer:

3cm

Step-by-step explanation:

=27cm2(area)÷3cm(length)

=9cm(width)

Degger [83]3 years ago

4 0

Formula for the area of a rectangle: A = length x width

Plug in the values we know into the formula: 27 = 9 x width

Divide both sides by 9: width = 3

The width of the rectangle is 3 cm.

Hope this helps!! :)

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g An urn contains 150 white balls and 50 black balls. Four balls are drawn at random one at a time. Determine the probability th

xxTIMURxx [149]

Answer:

With replacement, 0.2109 = 21.09% probability that there are 2 black balls and 2 white balls in the sample.

Without replacement, 0.2116 = 21.16% probability that there are 2 black balls and 2 white balls in the sample.

Step-by-step explanation:

For sampling with replacement, we use the binomial distribution. Without replacement, we use the hypergeometric distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Sampling with replacement:

I consider a success choosing a black ball, so $p = \frac{50}{150+50} = \frac{50}{200} = 0.25$

We want 2 black balls and 2 white, 2 + 2 = 4, so $n = 4$ , and we want P(X = 2).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{4,2}.(0.25)^{2}.(0.75)^{2} = 0.2109$

With replacement, 0.2109 = 21.09% probability that there are 2 black balls and 2 white balls in the sample.

Sampling without replacement:

150 + 50 = 200 total balls, so $N = 200$

Sample of 4, so $n = 4$

50 are black, so $k = 50$

We want P(X = 2).

$P(X = 2) = h(2,200,4,50) = \frac{C_{50,2}*C_{150,2}}{C_{200,4}} = 0.2116$

Without replacement, 0.2116 = 21.16% probability that there are 2 black balls and 2 white balls in the sample.

3 0

3 years ago

If we divide the polynomial p by the factor x − c and we obtain the equation p(x) = (x − c)q(x) + r(x), then we say that x − c i

Schach [20]

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6 0

3 years ago

What is another way to say 14 over 16 in fractions ( preferably in lowest number example 4 over 8 equals 1 over 2 both are halve

Degger [83]

Answer:

7/8

Step-by-step explanation:

The greatest common factor in each is two so you divide both the numerator and the denominator by two to simplify the fraction

4 0

3 years ago

Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side

lorasvet [3.4K]

Answer:

Step-by-step explanation:

cot x sec⁴ x = cot x+2 tan x +tan³x

L.H.S = cot x sec⁴x

=cot x (sec²x)²

=cot x (1+tan²x)² [ ∵ sec²x=1+tan²x]

= cot x(1+ 2 tan²x +tan⁴x)

=cot x+ 2 cot x tan²x+cot x tan⁴x

=cot x +2 tan x + tan³x [ ∵cot x tan x $=\frac{ \textrm{tan x }}{\textrm{tan x}}$ =1]

=R.H.S

(sin x)(tan x cos x - cot x cos x)=1-2 cos²x

L.H.S =(sin x)(tan x cos x - cot x cos x)

= sin x tan x cos x - sin x cot x cos x

$=\textrm{sin x cos x }\times\frac{\textrm{sin x}}{\textrm{cos x} } - \textrm{sinx}\times\frac{\textrm{cos x}}{\textrm{sin x}}\times \textrm{cos x}$

= sin²x -cos²x

=1-cos²x-cos²x

=1-2 cos²x

=R.H.S

1+ sec²x sin²x =sec²x

L.H.S =1+ sec²x sin²x

= $1+\frac{{sin^2x}}{cos^2x}$ [ $\textrm{sec x}=\frac{1}{\textrm{cos x}}$ ]

=1+tan²x $[\frac{\textrm{sin x}}{\textrm{cos x}} = \textrm{tan x}]$

=sec²x

=R.H.S

$\frac{\textrm{sinx}}{\textrm{1-cos x}} +\frac{\textrm{sinx}}{\textrm{1+cos x}} = \textrm{2 csc x}$

L.H.S= $\frac{\textrm{sinx}}{\textrm{1-cos x}} +\frac{\textrm{sinx}}{\textrm{1+cos x}}$

$=\frac{\textrm{sinx(1+cos x)+{\textrm{sinx(1-cos x)}}}}{\textrm{(1-cos x)\textrm{(1+cos x})}}$

$=\frac{\textrm{sinx+sin xcos x+{\textrm{sinx-sin xcos x}}}}{{(1-cos ^2x)}}$

$=\frac{\textrm{2sin x}}{sin^2 x}$

= 2 csc x

= R.H.S

-tan²x + sec²x=1

L.H.S=-tan²x + sec²x

= sec²x-tan²x

= $\frac{1}{cos^2x} -\frac{sin^2x}{cos^2x}$

$=\frac{1- sin^2x}{cos^2x}$

$=\frac{cos^2x}{cos^2x}$

8 0

4 years ago

Helppp please this is urgent :) thank you

larisa [96]

The slope would be -5/4 or -1.25 because the y-intercept is (0,5) and x-intercept is (4,0), thus giving y = 5 - (5 x)\/4

6 0

4 years ago