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3 years ago

2443x228x67x890x103x61x8*1

Mathematics

1 answer:

lukranit [14]3 years ago

4 0

Answer:

1032x176x67x890x30x6x8

1032×176×67×890×30×6×8x^6

1.56^13×x^6

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A triangle has side lengths of (q+r)(q+r) centimeters, (5q-10s)(5q−10s) centimeters, and (5s-7r)(5s−7r) centimeters. Which expre

Stolb23 [73]

The expression represents the perimeter, in centimeters, of the triangle is 6q - 6r - 5s

<h3>What is the perimeter?</h3>

The formula for perimeter of a triangle is expressed as;

Perimeter = a + b + c

Where a , b and c are the lengths of its side

(q+r)
(5q-10s)
(5s-7r)

Now, let's substitute the values

Perimeter = (q + r) + (5q - 10s) + (5s - 7r)

expand the bracket

Perimeter = q + r + 5q - 10s + 5s - 7r

collect like terms

Perimeter = q + 5q + r - 7r -10s + 5s

Add like terms

Perimeter = 6q - 6r - 5s

Thus, the expression represents the perimeter, in centimeters, of the triangle is 6q - 6r - 5s

Learn more about perimeter here:

brainly.com/question/24571594

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8 0

1 year ago

Solve for b. 11 3 b b = [?] Pythagorean Theorem: a2 + b2 = c2 Enter.

balu736 [363]

Answer:

√112

Step-by-step explanation:

a^2+b^2=c^2

3^2+b^2=11^2

9+b^2=121

b^2=121-9

b^2=112

b=√112

Mark me as brainliest

3 0

3 years ago

Until 2014, the mass of a kilogram was defined as exactly equal to the mass of a special

emmainna [20.7K]

Step-by-step explanation:

We have,

Diameter of cylinder, d = 39.2 mm

Radius, r = 19.6 cm

Height of the cylinder, h = 39.2 mm

It is required to find the surface area of this cylinder. The formula of the surface area of cylinder is given by :

$S=2\pi rh+2\pi r^2\\\\S=2\pi r(h+r)$

Putting all values we get :

$S=2\times\dfrac{22}{7}\times19.6\cdot(39.2+19.6)\\\\S=7244.16\ mm^2$

7 0

3 years ago

A manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective

Inessa05 [86]

Answer:

(a) P(X $\leq$ 20) = 0.9319

(b) Expected number of defective light bulbs = 15

Step-by-step explanation:

We are given that a manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective independently. A box of 150 bulbs is selected at random.

Firstly, the above situation can be represented through binomial distribution, i.e.;

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

where, n = number of samples taken = 150

r = number of success

p = probability of success which in our question is % of bulbs that

are defective, i.e. 10%

<em>Now, we can't calculate the required probability using binomial distribution because here n is very large(n > 30), so we will convert this distribution into normal distribution using continuity correction.</em>

So, Let X = No. of defective bulbs in a box

<u>Mean of X</u>, $\mu$ = $n \times p$ = $150 \times 0.10$ = 15

<u>Standard deviation of X</u>, $\sigma$ = $\sqrt{np(1-p)}$ = $\sqrt{150 \times 0.10 \times (1-0.10)}$ = 3.7

So, X ~ N( $\mu = 15, \sigma^{2} = 3.7^{2})$

Now, the z score probability distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

(a) Probability that this box will contain at most 20 defective light bulbs is given by = P(X $\leq$ 20) = P(X < 20.5) ---- using continuity correction

P(X < 20.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{20.5-15}{3.7}$ ) = P(Z < 1.49) = 0.9319

(b) Expected number of defective light bulbs found in such boxes, on average is given by = E(X) = $n \times p$ = $150 \times 0.10$ = 15.

5 0

3 years ago

(3f–3)^2 pls help<br> aaaaaaaaaaaaaaaaaaaa

julia-pushkina [17]

9f^2-18f+9 is the answer

6 0

3 years ago

24*43x22*8x67x890x10*3x6*1x8*1

2443x228x67x890x103x61x8*1