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4 years ago

I need to know answer

Mathematics

1 answer:

Crank4 years ago

7 0

what answer are you talking about you don't know what the question is we don't know what the question is what are you asking do you know the question

You might be interested in

–24 + 12d = 2(d – 3) + 22 what does e=

Kitty [74]

Answer:

d = 4

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

$-24 + 12d = 2(d-3) + 22$
$-24 + 12d = (2)(d) + (2)(-3) + 22$
$-24 + 12d = 2d -6 + 22$
$12d - 24 = 2d + 16$

Step 2: Subtract 2d from both sides.

$12d - 24 - 2d = 2d + 16 - 2d$
$10d - 24 = 16$

Step 3: Add 24 to both sides.

$10d - 24 + 24 = 16 + 24$
$10d = 40$

Step 4: Divide both sides by 10.

$\frac{10d}{10} = \frac{40}{10}$
$d = 4$

Therefore, d = 4! I hope this helped you.

6 0

3 years ago

Read 2 more answers

The base of a triangle is 5 meters greater than the height. If the area of the triangle is 102 m2, which are the dimensions of t

kakasveta [241]

Answer: The base is 17 meters long, and the height is 12 meters.

Step-by-step explanation:

The area of a triangle is:

A = b*h/2

where:

b = base

h = height.

For this particular triangle, we have that:

b = 5m + h

A = 102 m^2 = b*h/2

We can replace the first equation into the second one:

b*h/2 = 102 m^2

(5m + h)*h/2 = 102 m^2

(5m + h)*h = 2*102 m^2 = 204m^2

5m*h + h^2 = 204m^2

We can rewrite this as a quadratic equation:

h^2 + 5m*h - 204m^2 = 0

The solutions are given by the Bhaskara formula, the solutions are:

$h = \frac{-5m +- \sqrt{(+5m)^2 -4*1*(-204m^2)} }{2*1} = \frac{-5m+-29m}{2}$

Then the two solutions are:

h = (-5m - 29m)/2 = -17 m

This is a negative height, that is not really defined, so this will be discarded.

The other solution is:

h = (-5m + 29m)/2 = 12m

This is positive, so this is the option we will use.

Knowing h, we can find the value of b.

b = 5m + h = 5m + 12m = 17m

Then the area is:

A = 12m*17m/2 = 102m^2

5 0

3 years ago

What is the IQR (Interquartile Range) for the following data? 20, 27, 28, 29,

Flauer [41]

Answer:

IQR= 9

Q3-Q1

37-28=9 IQR

6 0

3 years ago

Experience shows that a ski lodge will be full (153 guests) if there is a heavy snow fall while only partially full (62 guests)

Leokris [45]

<h2>Answer:</h2><h2>The expected number of guests if the probability for a heavy snow fall is .40 = 98.4 </h2>

Step-by-step explanation:

The total number of guests for lodge to be full = 153

During heavy snow fall, the lodge will be full.

The expected number of guests during heavy snow fall = 153

During partial snow fall, the expected number of guests = 62

The probability of heavy snowfall = 0.4

The probability of partial snow fall = 1 - 0.4 = 0.6

The expected number of guests = [(0.4) (153)] + [(0.6) (62)]

= 61.2 + 37.2

= 98.4

6 0

3 years ago

1.Suppose that scores on a knowledge test are normally distributed with a mean of 71 and a standard deviation of 6

Degger [83]

Answer:

a) For this case we can see the distribution in the figure attached is a bell shaped graph and symmetrical around 71

b) $z = \frac{76-71}{6}= 0.833$

We can see the value of 76 labeled in the second picture attached

c) $P(X>76)$

And using the z score we have this using the normal standard table or excel:

$P(Z>0.833) = 1-P(Z$

d) $n = 185*0.202= 154.16$

We can say that about 154 and 155 students scored higher than Angelica

Step-by-step explanation:

We know that X represent the random variable scores of knowledge test and is given by:

$X \sim N (\mu = 71, \sigma =6)$

Part a

For this case we can see the distribution in the figure attached is a bell shaped graph and symmetrical around 71

Part b

For this case the z score is given by:

$z = \frac{X- \mu}{\sigma}$

And replacing we got:

$z = \frac{76-71}{6}= 0.833$

We can see the value of 76 labeled in the second picture attached

Part c

We want this probability:

$P(X>76)$

And using the z score we have this using the normal standard table or excel:

$P(Z>0.833) = 1-P(Z$

Part d

For this case we can find the number desired like this:

$n = 185*0.202= 154.16$

We can say that about 154 and 155 students scored higher than Angelica

6 0

4 years ago