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3 years ago

NEED HELP ASAP!!

Mathematics

1 answer:

lesya [120]3 years ago

6 0

B)50.97....................

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if one card is drawn from a regular deck of cards, what is the probability the card will not be an ace?

satela [25.4K]

There are 52 cards in a deck of cards so that makes 52 your denominator. There are 4 aces in a deck of cards, so 52 - 4 = 48 making 48 your numerator. so the probability the card won't be an ace is 48 out of 52.

48/52 chance that the card will not be an ace.

Hope this helps you. :-)

4 0

3 years ago

Read 2 more answers

What is the 25th term given a₁ = 8 and r = 2?

kondaur [170]

Answer:

134217728

Step-by-step explanation:

The geometric sequence formula is:

$\displaystyle{a_n = a_1r^{n-1}}$

We know that $\displaystyle{a_1=8}$ and $\displaystyle{r=2}$ , we also want to find the 25th term, substitute n = 25:

$\displaystyle{a_{25}=8\cdot 2^{25-1}}\\\\\displaystyle{a_{25}=8\cdot 2^{24}}\\\\\displaystyle{a_{25}=134217728}$

Therefore, the 25th term of sequence is 134217728

3 0

2 years ago

Plz plz heelp me i need heelp

frosja888 [35]

C) 1 makes the most sense

6 0

3 years ago

Read 2 more answers

16 emails to 6 text messages- 10 emails to 4 text messages proportional or not

Pavel [41]

Given:

16 emails to 6 text messages- 10 emails to 4 text messages

To find:

Whether 16 emails to 6 text messages- 10 emails to 4 text messages proportional or not.

Solution:

Check the ratio of emails to text. If the ratios are equal then the relation is proportional.

16 emails to 6 text messages.

$\dfrac{Emails}{Text}=\dfrac{16}{6}$

$\dfrac{Emails}{Text}=\dfrac{8}{3}$

$\dfrac{Emails}{Text}=8:3$

10 emails to 4 text messages.

$\dfrac{Emails}{Text}=\dfrac{10}{4}$

$\dfrac{Emails}{Text}=\dfrac{5}{2}$

$\dfrac{Emails}{Text}=5:2$

Since $\dfrac{8}{3}\neq \dfrac{5}{2}$ or $8:3\neq 5:2$ , therefore, it is not proportional.

5 0

3 years ago

Uppose y(t)=9e−5t

adoni [48]

If $y(t)=9e^{-5t}$ satisfies the ODE $y'-ky=0$ , then

$y'(t)=-45e^{-5t}$

and substituting into the ODE gives

$-45e^{-5t}-9ke^{-5t}=0\implies k=-5$

so that the ODE is $y'+5y=0$ .

I'm assuming that you're also supposed to determine the initial value $y_0$ given this ODE. You have

$y'+5y=0\implies e^{5t}y'+5e^{5t}y=0\implies(e^{5t}y)'=0\implies e^{5t}y=C\implies y=Ce^{-5t}$

Given $y(0)=y_0$ , you have

$y_0=C$

but since you already know that $y=9e^{-5t}$ , it follows that $C=9$ , and in turn that $y_0=9$ .

5 0

3 years ago