Can someone plz help?

1.) Some banks charge a fee on savings accounts that are left inactive for an extended period of time. The equation y = 5000 * (

oee [108]

Answer:

a. The initial amount of money that was left in this savings account was of 5000.

b. Decrease of 2% each year.

Step-by-step explanation:

Exponential function:

An exponential function, with an initial value of A(0), and a decay rate of r, as a decimal, is given by:

$A(t) = A(0)(1-r)^t$

In this question, we have:

$y = 5000*(0.98)^x$

a. What was the initial amount of money that was left in this savings account?

This is y(0) = 5000

The initial amount of money that was left in this savings account was of 5000.

b. What is the percent of change each year in this savings account?

First as a decimal.

1 - r = 0.98

r = 1 - 0.98

r = 0.02

So a decrease of 2% each year.

7 0

2 years ago

What is the value of x in the figure below? A-9 B-14 1/3 C-45 2/3 D-81

laila [671]

Answer:

A

Step-by-step explanation:

3x9+ 27 + 8= 35

8 0

2 years ago

Applying Properties of Exponents In Exercise, use the properties of exponents to simplify the expression.

NemiM [27]

Answer: (a) $e^3$

(b) $e^{10}$

(c) $\dfrac{1}{e^4}$

(d) $e^{4}$

Step-by-step explanation:

Properties of exponents :

1) $a^m\times a^n=a^{m+n}$

2) $\dfrac{a^m}{a^n}=a^{m-n}$

3) $(a^m)^n=a^{mn}$

4) $a^{-n}=\dfrac{1}{a^n}$

Now , we simplify the given expression using the above properties.

a) $(e^6)(e^{-3})$

$=e^{6+(-3)}$ [By property (1)]

$=e^{6-3}=e^3$ [∵ (+)(-)=(-)]

b) $(e^{-2})^{-5}$

$=e^{-2\times-5}$ [By property (3)]

$=e^{10}$ [∵ (-)(-)=(+)]

c) $(\dfrac{e^6}{e^2})^{-1}$

$=(e^{6-2})^{-1}$ [By property (2)]

$=(e^{4})^{-1}$

$=\dfrac{1}{e^4}$ [By property (4)]

d) $(e^3)^{\frac{4}{3}}$

$=e^{3\times\frac{4}{3}}$ [By property (3)]

$=e^{4}$

7 0

3 years ago

A multiple-choice examination has 20 questions, each with five possible answers, only one of which is correct. Suppose that one

kari74 [83]

Answer:

The probability that the student answers at least seventeen questions correctly is $8.03\times 10^{-10}$ .

Step-by-step explanation:

Let the random variable X represent the number of correctly answered questions.

It is provided all the questions have five options with only one correct option.

Then the probability of selecting the correct option is,

$P(X)=p=\frac{1}{5}=0.20$

There are n = 20 question in the exam.

It is also provided that a student taking the examination answers each of the questions with an independent random guess.

Then the random variable can be modeled by the Binomial distribution with parameters n = 20 and p = 0.20.

The probability mass function of X is:

$P(X=x)={20\choose x}\ 0.20^{x}\ (1-0.20)^{20-x};\ x =0,1,2,3...$

Compute the probability that the student answers at least seventeen questions correctly as follows:

$P(X\geq 17)=P (X=17)+P (X=18)+P (X=19)+P (X=20)$

$=\sum\limits^{20}_{x=17}{{20\choose x}\ 0.20^{x}\ (1-0.20)^{20-x}}\\\\=0.00000000077+0.000000000032+0.00000000000084+0.000000000000042\\\\=0.000000000802882\\\\=8.03\times10^{-10}$

Thus, the probability that the student answers at least seventeen questions correctly is $8.03\times 10^{-10}$ .

4 0

3 years ago

Part B: The next day the family ate 1/2 of what was left over after Mr.Swanson ate his late night snack.How much pizza did the f

Molodets [167]

what is part A? what is the main question?

6 0

2 years ago