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3 years ago

If EFG ≅ △ ABD so FG is = .......? help asap

Mathematics

2 answers:

Igoryamba3 years ago

5 0

Answer:

FG=BD

Step-by-step explanation:

hope this helps!!

RideAnS [48]3 years ago

4 0

It should be BD but sorry if i’m wrong

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The starting salaries of individuals with an MBA degree are normally distributed with a mean of $40,000 and a standard deviation

antiseptic1488 [7]

Answer:

$P(X>30000)=P(\frac{X-\mu}{\sigma}>\frac{30000-\mu}{\sigma})=P(Z-2)$

And we can find this probability using the complement rule and the normal standard distribution

$P(z>-2)=1-P(z$

Step-by-step explanation:

Let X the random variable that represent the starting salaries of individuals with a MBA of a population, and for this case we know the distribution for X is given by:

$X \sim N(40000,5000)$

Where $\mu=40000$ and $\sigma=5000$

We are interested on this probability

$P(X\geq 30000)$

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

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

Using this formula we got:

$P(X>30000)=P(\frac{X-\mu}{\sigma}>\frac{30000-\mu}{\sigma})=P(Z-2)$

And we can find this probability using the complement rule and the normal standard distribution

$P(z>-2)=1-P(z$

6 0

3 years ago

stack of mail consists of 8 bills, 10 letters, and 6 advertisements. One piece of mail is drawn at random and put aside. Then a

kozerog [31]

INFORMATION:

We know that:

- stack of mail consists of 8 bills, 10 letters, and 6 advertisements.

- One piece of mail is drawn at random and put aside. Then a second piece of mail is drawn.

And we must find P (both are letters)

STEP BY STEP EXPLANATION:

To find the probability, we need to know that we have two events. First, when one piece of mail is drawn at random and put aside and, second, when a second piece of mail is drawn.

These two events are dependent. If A and B are dependent events, P(A and B) = P(A) • P(B after A) where P(B after A) is the probability that B occurs after A has occurred.

So, first

- Probability of A (the first piece is letter)

$P(A)=\frac{favorable\text{ }cases}{total\text{ cases}}=\frac{10}{24}$

- Probability of B after A

Since A already occurred and one piece of the mail was drawn (a letter), now in total we would have 9 letter and 23 total pieces

$P(B\text{ after }A)=\frac{9}{23}$

Finally, replacing in the initial formula

$P(A\text{ and }B)=\frac{10}{24}\cdot\frac{9}{23}=\frac{90}{552}=0.1630$

Finally, the probability would be 0.1630

ANSWER:

P (both are letters) = 0.1630

8 0

1 year ago

Evaluate The following expression . in e^e

erik [133]

$\bf \textit{Logarithm Cancellation Rules} \\\\ \stackrel{\stackrel{\textit{let's use this one}}{\downarrow }}{log_a a^x = x}\qquad \qquad a^{log_a x}=x \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ln(e^e)\implies log_e(e^e)\implies e$

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3 years ago

Determine if the statement is always, sometimes or never true. Equilateral triangles are isosceles triangles.

tia_tia [17]

Never .. isoscels is unequal. I'm pretty sure

7 0

3 years ago

Please help, i forgot which was which... (positive or negative coefficient)

Vesna [10]

<h3>Answer: Negative</h3>

Reason:

The template $y = ax^2+bx+c$ applies to any quadratic to graph out a parabola. The coefficient for the x^2 term is 'a', and it solely determines whether the parabola opens upward or downward.

If 'a' is negative, then the parabola opens downward. The way to remember this is that 'a' being negative forms a negative frown.

On the other hand if 'a' is positive, then it forms a positive smile, and the parabola opens upward.

In this case, the points are fairly close to a parabola opening downward. This means 'a' is negative and a < 0.

8 0

2 years ago