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Arisa [49]
3 years ago
8

The weather forecast says that the probability of being cloudy tomorrow is 30% and the probability of raining is 25%. If it is c

loudy, the probability of it raining is 45%. what is the probability of it being cloudy and raining.
a. 7.5%
b. 15%
c. 13.5%
d. 11.25%​
Mathematics
1 answer:
Bumek [7]3 years ago
6 0
<h3>Answer:</h3>

\large\boxed{C).\,13.5\%}

<h3>Step-by-step explanation:</h3>

In this question, we're trying to find the probability of it being cloudy and raining.

In this case, we know that:

  • Probability of it being cloudy is 30%
  • Probability of it raining is 25% (this is necessarily not needed)
  • If it's cloud, the probability of it raining is 45%

With the information above, we can find the probability.

We know that from a 100% scale, the chance of it being cloudy is 30%.

We know that if it's cloudy, the chances of raining is 45%

To find the probability of it being cloudy and raining, we would multiply 0.3 (30%) by 0.45 (45%)

Solve:

0.3*0.45=0.135\\\\\text{Move the decimal place two times to the right}\\\text{to turn it into a percent}\\\\\boxed{13.5\%}

Your answer would be C). 13.5%

<h3>I hope this helped you out.</h3><h3>Good luck on your academics.</h3><h3>Have a fantastic day!</h3>
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12/18 ATB - Review question -<br> Evaluate 18x + 75 when x= 8.
boyakko [2]

Answer:

solution

x=8

so,

18 .8 +75

144 +75

219

7 0
3 years ago
A professor finds that the average SAT score among all students attending his college is 1150 ± 150 (μ ± σ). He polls his class
solniwko [45]

Answer:

Step-by-step explanation

Hello!

Be X: SAT scores of students attending college.

The population mean is μ= 1150 and the standard deviation σ= 150

The teacher takes a sample of 25 students of his class, the resulting sample mean is 1200.

If the professor wants to test if the average SAT score is, as reported, 1150, the statistic hypotheses are:

H₀: μ = 1150

H₁: μ ≠ 1150

α: 0.05

Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~~N(0;1)

Z_{H_0}= \frac{1200-1150}{\frac{150}{\sqrt{25} } } = 1.67

The p-value for this test is  0.0949

Since the p-value is greater than the level of significance, the decision is to reject the null hypothesis. Then using a significance level of 5%, there is enough evidence to reject the null hypothesis, then the average SAT score of the college students is not 1150.

I hope it helps!

7 0
3 years ago
I need to solve for x, help please.
BaLLatris [955]

Answer:

x = 52   i hope this helps!   :)

Step-by-step explanation:

so first you gotta realize that every triangle has an interior angle measurement sum of 180 degrees

so we are given 3 angles   74 degrees  x + 2  and x

we now are going to put this into an equation   74 + x + 2 + x = 180

now lets combine like terms to get   76 + 2x = 180

then we are going to subtract the 76 from both sides   2x = 104

now lets divide both sides by 2   x = 52

so x = 52

we can plug in the value of x into both of the other equations to find the angle measurement

52 degrees

52 + 2 = 54 degrees

and then we have 74 degrees

add the 52, 54, and the 74 to make sure it is equal to 180 degrees

106 + 74 = 180

so yes, x does indeed equal 52 degrees

3 0
3 years ago
Read 2 more answers
Pepe tenía dinero ahorrado en su alcancia pero le retiró $75 para comprar un jugete. Ahora tiene $250 en su alcancia ¿Cuánto din
gregori [183]
El tenia $325

$250+$75=$325

Espero q te ayude :)
4 0
3 years ago
Here are the endpoints of the segments BC, FG, and JK.<br> B, −67
yulyashka [42]

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F(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-4})\qquad G(\stackrel{x_2}{1}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ FG=\sqrt{[1 - (-2)]^2 + [-2 - (-4)]^2}\implies FG=\sqrt{(1+2)^2+(-2+4)^2} \\\\\\ FG=\sqrt{9+4}\implies \boxed{FG=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ J(\stackrel{x_1}{4}~,~\stackrel{y_1}{2})\qquad K(\stackrel{x_2}{5}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}

JK=\sqrt{[5 - 4]^2 + [-2 - 2]^2}\implies JK=\sqrt{1^2+(-4)^2}\implies \boxed{JK=\sqrt{17}} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \overline{BC}\cong \overline{FG}~\hfill

4 0
2 years ago
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