Find the area of the shape below

Mathematics

1 answer:

lilavasa [31]3 years ago

7 0

Area= LxW so u take 14+9=23 then 20+11=31 then multiple the length and the width then your answer will be 713cm

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The Siberian tiger girl 10 four over 5 feet long express the length as a decimal

zalisa [80]

The decimal answer would be 10.8
Hope it helped you!!

6 0

4 years ago

Assume that blood pressure readings are normally distributed with a mean of 123 and a standard deviation of 9.6. If 144 people a

Vanyuwa [196]

Answer:

The probability is 0.9938

Step-by-step explanation:

In this question, we are asked to calculate the probability that the mean blood pressure readings of a group of people is less than a certain reading.

To calculate this, we use the z score.

Mathematically;

z = (mean - value)/(standard deviation/√N)

From the question, we can identify that the mean is 125, the value is 123 , the standard deviation is 9.6 and N ( total population is 144)

Let’s plug these values;

z = (125-123)/(9.6/√144) = 2.5

Now we proceed to calculate the probability with a s score less than 2.5 using statistical tables

P(z<2.5) = 0.9938

8 0

3 years ago

Harold uses the binomial theorem to expand the binomial (3x^5 - 1/9y^3)^4

riadik2000 [5.3K]

<h3>The complete question:</h3>

Harold uses the binomial theorem to expand the binomial $(3x^5 -\dfrac{1}{9}y^3)^4$

(a) What is the sum in summation notation that he uses to express the expansion?

(b) Write the simplified terms of the expansion.

Answer:

(a). $(3x^5 -\dfrac{1}{9}y^3)^4=$$\sum_{k=0}^{n} \binom{4}{k}(3x^5)^{4-k}( -\dfrac{1}{9}y^3)^k $$$

(b). $(3x^5 -\dfrac{1}{9}y^3)^4=81x^{20}-12x^{15}y^3+\dfrac{2x^{10}y^6}{3}-\dfrac{4x^5y^9}{243}+\frac{y^{12}}{6561}$

Step-by-step explanation:

(a).

The binomial theorem says

$(x+y)^n=$$\sum_{k=0}^{n} \binom{n}{k}x^{n-k}y^k $$$

For our binomial this gives

$\boxed{(3x^5 -\dfrac{1}{9}y^3)^4=$$\sum_{k=0}^{n} \binom{4}{k}x^{4-k}y^k $$}$

(b).

We simplify the terms of the expansion and get:

$$$\sum_{k=0}^{n} \binom{4}{k}(3x^5)^{4-k}y^k $$= \binom{4}{0}(3x^5)^{4-0}(-\dfrac{1}{9}y^3 )^0+\binom{4}{1}(3x^5)^{4-1}(-\dfrac{1}{9}y^3 )^1+\\\\\binom{4}{2}(3x^5)^{4-2}(-\dfrac{1}{9}y^3 )^2+\binom{4}{3}(3x^5)^{4-3}(-\dfrac{1}{9}y^3 )^3+\binom{4}{4}(3x^5)^{4-4}(-\dfrac{1}{9}y^3 )^4$

$$$\sum_{k=0}^{n} \binom{4}{k}(3x^5)^{4-k}(-\frac{1}{9}y^3 )^k $$= (3x^5)^{4}+4(3x^5)^{3}(-\frac{1}{9}y^3 )+6(3x^5)^{2}(-\frac{1}{9}y^3 )^2+\\\\4(3x^5)(-\frac{1}{9}y^3 )^3+(-\frac{1}{9}y^3 )^4$