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3 years ago

Help me please of this question

Mathematics

1 answer:

kati45 [8]3 years ago

5 0

Answer:

Step-by-step explanation:

all adds up to 180 degrees

so 4x+3+2x+9=180

=6x+12=180

6x=168

x=28

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Evaluate the determinant for the following matrix [3 -5 / 1 1]

Natalka [10]

Answer:

Step-by-step explanation:

$\det\left[\begin{array}{ccc}a&b\\c&d\end{array}\right] =ad-bc\\\\\det\left[\begin{array}{ccc}3&-5\\1&1\end{array}\right] =(3)(1)+(-5)(1)=3-5=-2$

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3 years ago

You are baking cookies for your class. There are 23 total students in your class and you have baked 12 cookies. Write and solve

KonstantinChe [14]

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3 years ago

Is my answer correct

Vlad [161]

No because its explain how much cumulative tip total on day 5 so the answer is D.400

8 0

3 years ago

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponenti

melomori [17]

Answer:

a) 0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

b) Capacity of 252.6 cubic feet per second

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second).

This means that $m = 100, \mu = \frac{1}{100} = 0.01$

(a) Find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)

We have that:

$P(X > x) = e^{-\mu x}$

This is P(X > 190). So

$P(X > 190) = e^{-0.01*190} = 0.1496$

0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

(b) What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.08?

This is x for which:

$P(X > x) = 0.08$