Is -4x binomial trinomial or monomial

Mathematics

1 answer:

kenny6666 [7]3 years ago

8 0

The answer to your question is Monomial (:

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Vikki [24]

Answer:

5/6

Step-by-step explanation:

9/30 + 16/30 = 25/30 = 5/6

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3 years ago

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I need help with DE Precalculus, mainly trig identities. Would anyone be willing to zoom or text me to help tutor? Thanks!

liq [111]

Answer:

I would help but I'm really tired

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3 years ago

Find the area of triangle ABC. The figure is not drawn to scale.

balu736 [363]

Sin(63)=x/7
0.89100652418= x/7 > 0.89100652418 (7) = x/7(7)
6.23704566932=x This is the height
the area of a triangle is A=BH/2 or Base * Height /2
A= (6.23704566932)(8)/2
A=24.9481826773 or 24.95 cm^2

6 0

3 years ago

X–2 1/2=6 3/4. X=? x – 3.23 = 0.77. X=?

Over [174]

Answer:

x= 105/4

x=4

Step-by-step explanation:

5 0

3 years ago

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Average precipitation for the first 7 months of the year, the average precipitation in toledo, ohio, is 19.32 inches. if the ave

Colt1911 [192]

Part A:

The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\sigma}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

$P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\ \\ =1-P(z\ \textless \ -0.5410) \\ \\ =1-0.29426=\bold{0.7057}$

Part B:

The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that 5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

 $P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}$

7 0

3 years ago