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CaHeK987 [17]
2 years ago
15

Which properties are present in a table that represents an exponential function in the form y-b* when b > 1?

Mathematics
2 answers:
Oksana_A [137]2 years ago
7 0

Answer:

<u>Properties that are present are </u>

Property I

Property IV

Step-by-step explanation:

The function given is  y=b^x  where b > 1

Let's take a function, for example,  y=2^x

Let's check the conditions:

I. As the x-values increase, the y-values increase.

Let's put some values:

y = 2 ^ 1

y = 2

and

y = 2 ^ 2

y = 4

So this is TRUE.

II. The point (1,0) exists in the table.

Let's put 1 into x and see if it gives us 0

y = 2 ^ 1

y = 2

So this is FALSE.

III. As the x-value increase, the y-value decrease.

We have already seen that as x increase, y also increase in part I.

So this is FALSE.

IV. as the x value decrease the y values decrease approaching a singular value.

THe exponential function of this form NEVER goes to 0 and is NEVER negative. So as x decreases, y also decrease and approached a value (that is 0) but never becomes 0.

This is TRUE.

Option I and Option IV are true.

lyudmila [28]2 years ago
3 0

Answer:I and IV

Step-by-step explanation:

You might be interested in
For a display, identical cubic boxes are stacked in square layers. Each layer consists of cubic boxes arranged in rows that form
serg [7]

Answer:

285  boxes are in the display

Step-by-step explanation:

Given data

top layer box = 1

last row box = 81

to find out

how many box

solution

we know that every row is a square so that if the bottom layer has 81 squares it mean this is 9² and every row has one lesser box

so that next row will have 8^2 and than 7² and so on till 1²

so we can say that cubes in the rows as that

Sum of all Squares = 9² + 8² +..........+ 1²

Sum of Squares positive Consecutive Integers formula are

Sum of Squares of Consecutive Integers = (1/6)(n)(n+1)(2n+1)  

here n = 9 so equation will be

Sum of Squares of Consecutive Integers = (1/6) × (9) × (9+1) × (2×9+1)

Sum of Squares of Consecutive Integers = 285

so 285  boxes are in the display

7 0
2 years ago
Compare the values of the 2s and 5s in 55,220
bagirrra123 [75]
The five digit number, 55,220 contains to 5's and two 2's.  The 5's are in the ten thousand and one thousand columns. This means that one five represents 50,000 and the second represents 5, 000.  The two's are in the hundreds and tens columns meaning one 2 represents 200 and the other 2 represents 20. In a direct comparison of these numbers the 5's equal 55,000 and the 2's equal 220.
6 0
2 years ago
Prof. okamoto selects 3 different students to report on 3 different current events. if there are 27 students in the class, in ho
kenny6666 [7]
Consider this option:
C³₂₇=27!/(3!*24!)=25*13*9=2925 ways to select 3 students.
8 0
2 years ago
Haruka hiked several kilometers in the morning. She hiked only 6 kilometers in the afternoon, which was 25%, percent less than s
Semenov [28]

Answer:

14 km

Step-by-step explanation:

6 km in the afternoon

x km in the morning

6= x- 25%

6= 0.75x

x= 6/0.75

x= 8 km

total 6+8= 14 km

4 0
2 years ago
A random sample of 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of 12
marissa [1.9K]

Answer:

Step-by-step explanation:

Hello!

You have two random samples obtained from two different normal populations.

Sample 1

n₁= 15

X[bar]₁= 350

S₁= 12

Sample 2

n₂= 17

X[bar]₂= 342

S₂= 15

At α: 0.05 you need to obtain the p-value for testing variances for a one tailed test.

If the statistic hypotheses are:

H₀: σ₁² ≥ σ₂²

H₁: σ₁² < σ₂²

The statistic to test the variances ratio is the Stenecor's-F test. F_{H_0}=(\frac{S^2_1}{Sigma^2_1}) * (\frac{S^2_2}{Sigma^2_2} )~F_{n_1-1;n_2-1}

F_{H_0}= \frac{(12)^2}{(15)^2} * 1= 0.64

The p-value is:

P(F_{14;16}≤0.64)= 0.02

I hope it helps!

7 0
2 years ago
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