Can yall help answer this asap

What digit is in the ten-thousands place in the number 81,473

Ugo [173]

Answer:

8

Explanation:

8 - Ten-Thousands

1 - Thousands

4 - Hundreds

7 - Tens

3 - Ones

3 0

3 years ago

Read 2 more answers

The sun is about 10 • 10 million miles away from earth. Write 10 • 10 using an exponent. Then find the value of the power. How m

marissa [1.9K]

Answer:

so if 10 times 10 is 100
your answer should be 10 to the 2nd power ( the 2nd power is the little number that goes on the top right of 10)

3 0

3 years ago

How much is 5/9 of -3/5? A) 1/2 B) -1/3 C) 1/7 D) -25/27

irina1246 [14]

The answer is A 1/2 sorry if I’m wrong if not A its c

5 0

3 years ago

Read 2 more answers

Which is the perimeter of the preceding figure

OlgaM077 [116]

THE PERIMETER OF THE PRECEDING FIGURE 17.8 METERS

7 0

3 years ago

A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or

Tamiku [17]

Answer:

The students should request an examination with 5 examiners.

Step-by-step explanation:

Let X denote the event that the student has an “on” day, and let Y denote the

denote the event that he passes the examination. Then,

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

The events ( $Y|X$ ) follows a Binomial distribution with probability of success 0.80 and the events ( $Y|X^{c}$ ) follows a Binomial distribution with probability of success 0.40.

It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,

$P(X)=2\cdot P(X^{c})$

Then,

$P(X)+P(X^{c})=1$

⇒

$2P(X^{c})+P(X^{c})=1\\\\3P(X^{c})=1\\\\P(X^{c})=\frac{1}{3}$

Then,

$P(X)=1-P(X^{c})\\=1-\frac{1}{3}\\=\frac{2}{3}$

Compute the probability that the students passes if request an examination with 3 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}$

$=0.715$

The probability that the students passes if request an examination with 3 examiners is 0.715.

Compute the probability that the students passes if request an examination with 5 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}$

$=0.734$

The probability that the students passes if request an examination with 5 examiners is 0.734.

As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.

8 0

3 years ago

Can yall help answer this asap​

Can yall help answer this asap