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ivolga24 [154]
2 years ago
10

Consider f(x) = bx. Which statement(s) are true for 0 < b < 1? Check all that apply:

Mathematics
2 answers:
Margaret [11]2 years ago
4 0

Answer:

The Domain is all real numbers.

The range is y > 1.

The graph has a y-intercept of 1.

The function is always decreasing.

Step-by-step explanation:

Bess [88]2 years ago
3 0

we are given

f(x)=b^x

where 0<b<1

Domain:

domain is all possible values of x for which any function is defined

so, we can select any values of x for which function

it will be defined for all real x

so, domain is

(-\infty,\infty)

Range:

Range is all possible values of y for which x is defined

we are given that b is positive

so, value of function will always be positive

so, range is

(0,\infty)

or

y>0

x-intercept:

we can set f(x)=0

and then we can solve for x

f(x)=b^x=0

x is undefined

so, x-intercept does not exist

Increasing or decreasing:

Since, 0<b<1

so, b is positive value less than 1

so, as we increase value of , b^x will keep decreasing

so, this is decreasing

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Suppose that the exam scores for students in a large university course are normally distributed with an unknown mean and standar
HACTEHA [7]

Answer:

Critical value t-score=2.701.

Step-by-step explanation:

When we calculate a confidence interval with an unknown population standard deviation, we estimate it from the sample standard deviation and use the t-score instead of the z-score.

The critical value for t depends on the level of confidence and the degrees of freedom.

If the sample size is 42, the degrees of freedom are:

df=n-1=42-1=41

For a confidence level of 99% and 41 degrees of freedom, the critical value of t is t=2.701.

6 0
2 years ago
11) The Cost of maintaing a
dezoksy [38]

Answer:

a). Cost of 44 pupils = $14265

b). Least number of pupils = 31

Step-by-step explanation:

The given question is incomplete; here is the complete question.

The cost of maintaining a school is partly constant and partly varies as the number of pupils. With 50 pupils, the cost is $15,705.00 and with 40 pupils, it is $13,305.00.

(a) Find the cost when there are 44 pupils.

(b) If the fee per pupil is $360.00, what is the least number of pupils for which the school can run without a loss?

Let the equation representing the total cost of maintaining a school is,

C = ax + b

Where C = Total cost of maintaining a school

a = Fee per pupil

b = Fixed running cost

x = number of pupils

a). Cost of 50 pupils = $15705

    Equation will be,

    15705 = 50a + b -------(1)

    Cost of 40 pupils = $13305

    Equation will be,

    13305 = 40a + b --------(2)

    By subtracting equation (2) from equation (1),

    15705 - 13305 = (50a + b) - (40a + b)

    2400 = 10a

    a = 240

    From equation (1),

    b = 3705

    Equation representing the total cost will be,

    C = 240x + 3705

    If x = 44

    C = 240(44) + 3705

    C = $14265

b). If the fee per pupil 'a' = $360

    Let the number of pupils = p

    Total fee of 'p' pupils = $360p

    Total cost to run the school will be = 3705 + 240p

    For the school not to be in the loss,

    360p ≥ 3705 + 240p

    360p - 240p ≥ 3705

    120p ≥ 3705

    p ≥ \frac{3705}{120}

    p ≥ 30.875

    Therefore, to run the school without loss, number pupils should be at least 31.

3 0
2 years ago
Factorise 15x+25<br><br>thank you for any help​
scoray [572]

Answer: 5(3x+5)

Step-by-step explanation:

To factor this expression we need to pull out the greatest common factor of the 2 numbers, 15 and 25. This factor is 5.

Now divide both numbers by 5 and put a 5 on the outside of the new expression to look like: 5(3x+5) and this is as far as it can be factored.

4 0
2 years ago
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What is the difference of the two polynomials?<br><br> (7y2 + 6xy) – (–2xy + 3)
Paraphin [41]
<span>(7y2 + 6xy) – (–2xy + 3)

7y^2 + 6xy +2xy +3

7y^2 +8xy +3

8xy + 7y^2 +3</span>
4 0
2 years ago
Read 2 more answers
The diagram shows a circle inside a square.<br> a O<br> 16 cm<br> Work out the area of the circle.
LuckyWell [14K]

Answer:

perimeter of circle=2πr

16=2πr

r=16/2π=8/πcm

now.

area of circle=πr²=π(8/π)²=π×64/π²=64/(22/7)=64×7/22=20.36cm²

4 0
1 year ago
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