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3 years ago

What’s the area for each of these figures

Mathematics

1 answer:

Rainbow [258]3 years ago

8 0

Step-by-step explanation:

you should try multiplying all the sides and don't forget that a triangle is base times height and divided by two

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The width of a singles tennis court is 75% of the width of a doubles court. A doubles court Is 36 feet wide. How wide is a singl

yuradex [85]

Answer:

width of a singles court = 27 feet

Step-by-step explanation:

Width of a doubles tennis court = 36 feet

Width of a singles tennis court = 75% of 36

$width = 75\% \times 36 \\ width = \frac{75}{100} \times 36 \\ width = \frac{3} {4} \times 36 \\ width = \frac{108}{4} = 27$

8 0

3 years ago

Amir has a $2 off coupon for his favorite sub shop. To get the deal, he must buy a 6” sandwich and chips, a soda, or a cookie. H

Igoryamba

His total would be $5.50

5.50+0.75+1.25= 7.50

7.50-2(the coupon) = $5.50 total

3 0

3 years ago

Read 2 more answers

I need help on this one please

katrin2010 [14]

Answer:

Step-by-step explanation:

(-2,4), (0,6), and(2,8)

6 0

3 years ago

Given that f(x)=2x/3 evaluate f^-1(6)

abruzzese [7]

Answer:

Step-by-step explanation:

we know by definition that

$f^{-1}(f(x))=x\\\\f^{-1}(\frac{2x}{3})=x\\\\f^{-1}(x)=ax\\\\f^{-1}(\frac{2x}{3})=a(\frac{2x}{3})=x\\\\f^{-1}(x)=\frac{3x}{2}$

so now we evaluate

$f^{-1}(6)=\frac{3*6}{2}\\\\=3*3\\\\f^{-1}(6)=9$

and if we want to do an extra step

$f^{-1}(f(6))=6\\\\f^{-1}(\frac{2*6}{3})=f^{-1}(4)=\frac{3*4}{2}=6$

which works.

4 0

2 years ago

In 1898 L. J. Bortkiewicz published a book entitled The Law of Small Numbers. He used data collected over 20 years to show that

attashe74 [19]

Answer:

(a) The probability of more than one death in a corps in a year is 0.1252.

(b) The probability of no deaths in a corps over 7 years is 0.0130.

Step-by-step explanation:

Let <em>X</em> = number of soldiers killed by horse kicks in 1 year.

The random variable $X\sim Poisson(\lambda = 0.62)$ .

The probability function of a Poisson distribution is:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,...$

(a)

Compute the probability of more than one death in a corps in a year as follows:

P (X > 1) = 1 - P (X ≤ 1)

= 1 - P (X = 0) - P (X = 1)

$=1-\frac{e^{-0.62}(0.62)^{0}}{0!}-\frac{e^{-0.62}(0.62)^{1}}{1!}\\=1-0.54335-0.33144\\=0.12521\\\approx0.1252$

Thus, the probability of more than one death in a corps in a year is 0.1252.

(b)

The average deaths over 7 year period is: $\lambda=7\times0.62=4.34$

Compute the probability of no deaths in a corps over 7 years as follows:

$P(X=0)=\frac{e^{-4.34}(4.34)^{0}}{0!}=0.01304\approx0.0130$

Thus, the probability of no deaths in a corps over 7 years is 0.0130.

6 0

2 years ago