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2 years ago

Each angle of an equiangular triangle is what degree angle

1 answer:

5 0

An equiangular triangle adds up to 180 so divide that by 3, which is the number of sides, and you get, 60, therefore, each side is 60<span>° angle.</span>

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Multiply. Write your answer as a fraction in simplest form. −1/4×(−4/3)=

Sophie [7]

Answer:

$\frac{1}{3}$

Step-by-step explanation:

Given

- $\frac{1}{4}$ × - $\frac{4}{3}$ ( negative times negative = positive )

= $\frac{1}{4}$ × $\frac{4}{3}$ ( cancel the 4 on the numerator and denominator )

= $\frac{1}{3}$

5 0

2 years ago

The area of a triangle is comtinous or discrete

Colt1911 [192]

Answer:

Continuous

Step-by-step explanation:

The area of a triangle is continuous. It can be any measure of fraction or decimal depending on the side lengths. It has no restrictions to be only whole numbers or a specific set of numbers.

3 0

3 years ago

a full container of juice 64 fluid ounces.how many 7 fluid ounces serving of juice are in a full containet?

saul85 [17]

To find the amount of 7 fluid ounce servings in this container, divide 64 by 7 to get the answer of about 9.14 (rounded to the nearest one hundredth). So 9 full servings and an additional .142857... ounces as well.

Hope this helps!

5 0

2 years ago

Read 2 more answers

Which expression is equivalent to -18a^-2b^5/-12a^-4b^-6

crimeas [40]

Answer:

c. 3a^2b^11/2

Step-by-step explanation:

The applicable rule of exponents is ...

(a^b)/(a^c) = a^(b-c)

$\dfrac{-18a^{-2}b^5}{-12a^{-4}b^{-6}}=\dfrac{-18}{-12}a^{-2-(-4)}b^{5-(-6)}=\boxed{\dfrac{3a^2b^{11}}{2}}$

3 0

2 years ago

Suppose that X has an exponential distribution with mean equal to 10. Determine the following: a. P(X > 10) b. P(X > 20) c

GrogVix [38]

Answer:

(a) The value of P (X > 10) is 0.3679.

(b) The value of P (X > 20) is 0.1353.

(d) The value of x is 30.

Step-by-step explanation:

The probability density function of an exponential distribution is:

$f(x)=\lambda e^{-\lambda x};\ x>0, \lambda>0$

The value of E (X) is 10.

The parameter λ is:

$\lambda=\frac{1}{E(X)}=\frac{1}{10}=0.10$

(a)

Compute the value of P (X > 10) as follows:

$P(X>10)=\int\limits^{\infty}_{10} {0.10 e^{-0.10 x}} \, dx \\=0.10\int\limits^{\infty}_{10} { e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10} |^{\infty}_{10}\\=|e^{-0.10 x} |^{\infty}_{10}\\=e^{-0.10\times10}\\=0.3679$

Thus, the value of P (X > 10) is 0.3679.

(b)

Compute the value of P (X > 20) as follows:

$P(X>20)=\int\limits^{\infty}_{20} {0.10 e^{-0.10 x}} \, dx \\=0.10\int\limits^{\infty}_{20} { e^{-0.10 x}} \, dx\\=0.10|\frac{e^{-0.10 x}}{-0.10} |^{\infty}_{20}\\=|e^{-0.10 x} |^{\infty}_{20}\\=e^{-0.10\times20}\\=0.1353$