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Veronika [31]
3 years ago
15

eorge recorded these temperatures one day: Morning temperature = −2°F Afternoon temperature = +5°F Night temperature = 0°F Which

statement is correct, based on the given temperatures?
Mathematics
1 answer:
Dvinal [7]3 years ago
3 0
A. The Afternoon Temp. Was 5 Degrees above the Night Temp.
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The ratio of dogs to cats at a local animal shelter is 13/25. Which statement must be true?
Alinara [238K]

\bf \cfrac{dogs}{cats}\qquad \stackrel{ratio}{\cfrac{13}{25}}\qquad \cfrac{\leftarrow \textit{for every 13 dogs}}{\leftarrow \textit{there are 25 cats}}

4 0
3 years ago
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Rani takes a 40-question test. She answers 5% of the questions incorrectly
Feliz [49]

Answer:

2 Questions.

Step-by-step explanation:

To calculate a how much a percentage is of a number, you multiply the number by the percent.

So, we have the equation:

40 x 0.05 = 2

So, Rani got 2 questions incorrect.

Hope this helps!

5 0
2 years ago
Please help me this question
postnew [5]
Let's split it into rectange and tringle
BC= 2.1+ sqrt(3.2^2 - 1.9^2) = 4.67
7 0
2 years ago
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The figure above shows the probability density function for the random variable x. What is P(3≤x<7)?
qwelly [4]

The value of the probability P(3≤x<7) is 1

<h3>How to evaluate the probability expression?</h3>

The expression is given as: P(3≤x<7)

This is calculated using:

P(3 ≤ x < 7) = P(3) + P(4) + P(5) + P(6)

Using the figure of the probability density function (see attachment), we have:

P(3 ≤ x < 7) = 0.30 + 0.30 + 0.20 + 0.20

Evaluate

P(3 ≤ x < 7) = 1

Hence, the value of the expression is 1

Read more about probability density function at:

brainly.com/question/15318348

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4 0
1 year ago
Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth.
marishachu [46]

Answer:

Thus, the two root of the given quadratic equation  x^2-6=-x  is 2 and -3 .

Step-by-step explanation:

Consider, the given Quadratic equation, x^2-6=-x

This can be written as ,  x^2+x-6=0

We have to solve using quadratic formula,

For a given quadratic equation ax^2+bx+c=0 we can find roots using,

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}  ...........(1)

Where,  \sqrt{b^2-4ac} is the discriminant.

Here, a = 1 , b = 1 , c = -6

Substitute in (1) , we get,

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

\Rightarrow x=\frac{-(1)\pm\sqrt{(1)^2-4\cdot 1 \cdot (-6)}}{2 \cdot 1}

\Rightarrow x=\frac{-1\pm\sqrt{25}}{2}

\Rightarrow x=\frac{-1\pm 5}{2}

\Rightarrow x_1=\frac{-1+5}{2} and \Rightarrow x_2=\frac{-1-5}{2}

\Rightarrow x_1=\frac{4}{2} and \Rightarrow x_2=\frac{-6}{2}

\Rightarrow x_1=2 and \Rightarrow x_2=-3

Thus, the two root of the given quadratic equation x^2-6=-x is 2 and -3 .

7 0
3 years ago
Read 2 more answers
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