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puteri [66]
3 years ago
9

A = 1/2bh b= 12 h = 7 how to I go about getting this answer?

Mathematics
1 answer:
Finger [1]3 years ago
4 0
I know it is a triangle so 12×7=84÷2=42
You might be interested in
X + 2y = 7<br> 3x - 2y = -3
VladimirAG [237]
Use Subtraction

1. Subtract
x+2y=7
3x-2y=-3
-------------
4x+0y=4 SIMPLIFY 4x=4

2. DIVIDE

4x=4
x=1

Substitute that into either of the original equations to get x=1, y=3





4 0
3 years ago
3<br> (<br> x<br> +<br> 7<br> )<br> &lt;<br> 7<br> (<br> x<br> +<br> 2<br> )
Gre4nikov [31]
3(x+7)<7(x+2)

Inequality Form: x>7/4
Interval Notation: (7/4, infinity)
5 0
2 years ago
Autos de F 1. En un circuito de Australia, Hamilton hizo el récord de vuelta. Si lo normal a 240km/h para completar el circuito
olga2289 [7]

Answer:

Tenía que hacerlo en menos de 2 minutos a más de 240 km/h

Step-by-step explanation:

La velocidad normal dada para completar el circuito de 6.000 metros = 240 km/h

El tiempo normal que se tarda en completar el circuito de 6000 metros = 2 minutos

Dado que Hamilton estableció el récord de vuelta, el tiempo que tuvo que hacer para recorrer el récord de vuelta es de 6000 metros en menos de 2 minutos

La velocidad media con la que lo hizo es de más de 240 km/h.

6 0
2 years ago
Find the value of:<br> a) 5x when x = 6<br><br> B) 3y when y = -7
Y_Kistochka [10]

Answer:

a 30 b -21

Step-by-step explanation:

a 5x = 5(6)

= 30

b 3y = 3(-7)

= -21

7 0
3 years ago
Read 2 more answers
A college conducts a common test for all the students. For the Mathematics portion of this test, the scores are normally distrib
Jet001 [13]

Using the normal distribution, it is found that 58.97% of students would be expected to score between 400 and 590.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

Z = \frac{X - \mu}{\sigma}

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

\mu = 502, \sigma = 115

The proportion of students between 400 and 590 is the <u>p-value of Z when X = 590 subtracted by the p-value of Z when X = 400</u>, hence:

X = 590:

Z = \frac{X - \mu}{\sigma}

Z = \frac{590 - 502}{115}

Z = 0.76

Z = 0.76 has a p-value of 0.7764.

X = 400:

Z = \frac{X - \mu}{\sigma}

Z = \frac{400 - 502}{115}

Z = -0.89

Z = -0.89 has a p-value of 0.1867.

0.7764 - 0.1867 = 0.5897 = 58.97%.

58.97% of students would be expected to score between 400 and 590.

More can be learned about the normal distribution at brainly.com/question/27643290

#SPJ1

6 0
1 year ago
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