All categories

3 years ago

Use the diagram to determine which statement is true

Mathematics

2 answers:

hichkok12 [17]3 years ago

5 0

A. Area of ABCD - Area of DGA = Area of DEFG
s^2 - 1/2bh = s^2
(5)^2 - 1/2(4)(3) = (3)^2
25 - 1/2(12) = 9
25 - 24 = 9
1 not equal to 9

B. Area of ABCD - Area of GHIA = Area of DGA
s^2 - s^2 = 1/2bh
(5)^2 - (4)^2 = 1/2(4)(3)
25 - 16 = 1/2(12)
9 not equal to 6

C. Area of ABCD + Area of DGA = Area of GHIA
s^2 + 1/2bh = s^2
(5)^2 + 1/2(4)(3) = (4)^2
25 + 1/2(12) = 16
25 + 6 = 16
31 not equal to 16

D. Area of DEFG + Area of GHIA = Area of ABCD
s^2 + s^2 = s^2
(3)^2 + (4)^2 = (5)^2
9 + 16 = 25
25 = 25

The answer is D.

irina1246 [14]3 years ago

4 0

Answer:

Step-by-step explanation:

The DGA triangle is the right triangle.

We have three squares built on the sides of a given triangle.

Based on Pythagoras' theorem we have:

$A_{ABCD}=A_{DEFG}+A_{GHIA}$

You might be interested in

A consensus forecast is the average of a large number of individual analysts' forecasts. Suppose the individual forecasts for a

zaharov [31]

Answer:

a) $P(X>3.5)=P(\frac{X-\mu}{\sigma}>\frac{3.5-\mu}{\sigma})=P(Z>\frac{3.5-5}{1.2})=P(z>-1.25)$

And we can find this probability using the complement rule:

$P(z>-1.25)=1-P(z$

b) $P(X$

And we can find this probability uing the normal standard table:

$P(z$

c) $P(3.5$

And we can find this probability with this difference:

$P(-1.25$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

$P(-1.25$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(5,1.2)$

Where $\mu=5$ and $\sigma=1.2$

We are interested on this probability

$P(X>3.5)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

$P(X>3.5)=P(\frac{X-\mu}{\sigma}>\frac{3.5-\mu}{\sigma})=P(Z>\frac{3.5-5}{1.2})=P(z>-1.25)$

And we can find this probability using the complement rule:

$P(z>-1.25)=1-P(z$

Part b

We are interested on this probability

$P(X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

$P(X$

And we can find this probability uing the normal standard table:

$P(z$

Part c

$P(3.5$

And we can find this probability with this difference:

$P(-1.25$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

$P(-1.25$

3 0

3 years ago

The table below represents the atmospheric temperature at a location as a function of the altitude:

RSB [31]

from x15 to x 25 is 4 - -16 = -20 degree change

25 -15 = 10000 feet

-20/10 = -2 degrees every 1000 feet

7 0

3 years ago

I added pics cause it’s 2 in the morning.. I’m having back pain, thigh pain, cramps EVERYWHERE.. headache and just finished baby

goldfiish [28.3K]

Answer:

1/4^2 (The first one)

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Kira runs 4 miles in 30 minutes. at the same rate, how many miles would she run in 48 minutes?

NNADVOKAT [17]

Answer:

Step-by-step explanation:

4 0

3 years ago

Statistics question about random probability Cheese pastureized or Raw MilkA cheese can be classified as either raw-milk or past

blsea [12.9K]

Answer:

(a) Two cheeses are chosen at random. the probability that both cheeses are pasteurized is Pr(PP) = 0.82 x 0.82 = 0.6724 to 4 decimal places)

(b) Four cheeses are chosen at random. The probability that all four cheeses are pasteurized is Pr(PPPP) = 0.82 x 0.82 x 0.82 x 0.82 = 0.4521 to 4 decimal places

(c) What is the probability that at least one of four randomly selected cheeses is raw-milk is Pr(RPPP) Or Pr(RRPP) Or Pr(RRRP) Or Pr(RRRR)

= 0.1269 to 4 decimal places

It would not be unusual that at least one of four randomly selected cheeses is raw-milk, because the probability have a value between 0 and 1

Step-by-step explanation:

If is given that 80% of the cheese is classified as pasteurized.

It then implies that 20% of the cheese is classified as Raw-milk

Probability of pasteurized cheese is 0.82(Denoted by Pr(P))

Probability of raw-milk cheese is 0.18(Denoted as Pr(R))

(a) Two cheeses are chosen at random. the probability that both cheeses are pasteurized is Pr(PP) = 0.82 x 0.82 = 0.6724 to 4 decimal places)

(b) Four cheeses are chosen at random. The probability that all four cheeses are pasteurized is Pr(PPPP) = 0.82 x 0.82 x 0.82 x 0.82 = 0.4521 to 4 decimal places

(c) What is the probability that at least one of four randomly selected cheeses is raw-milk is Pr(RPPP) + Pr(RRPP) + Pr(RRRP) + Pr(RRRR)

(0.18 x 0.82 x 0.82 x 0.82) + (0.18 x 0.18 x 0.82 x 0.82) + (0.18 x 0.18 x 0.18 x 0.82) + (0.18 x 0.18 x 0.18 x 0.18) = 0.1269 to 4 decimal places

3 0

3 years ago