Y - 3 = -2(x -1) How do i do this?

Given that the domain is all real numbers, what is the limit of the range for the function ƒ(x) = 42x - 100?

slava [35]

A linear function has no restriction on range, unless there is one on the domain.

The range is all real numbers.

5 0

2 years ago

Is 3n a cubic a quadratic a linear or none of these

lilavasa [31]

Step-by-step explanation:

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Class 10

>>Maths

>>Polynomials

>>Revisiting Polynomials

>>Classify the following as linear, quadra

Question

Bookmark

Classify the following as linear, quadratic and cubic polynomials:

(i) x

2

+x

(ii) x−x

3

(Iii) y+y

2

+4

(iv) 1+x

(v) 3t

(vi) r

2

(vii) 7x

3

Medium

Solution

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(i) The highest degree of x

2

+x is 2, so it is a quadratic polynomials.

(ii) The highest degree of x−x

3

is 3, so it is a cubic polynomials.

(iii)The highest degree of y+y

2

+4 is 2, so it is a quadratic polynomials.

(iv) The highest degree of x in (1+x) is 1, so it is a linear polynomials.

(v)The highest degree of t in 3t is 1, so it is a linear polynomials.

(vi)The highest degree of r

2

is 2, so it is a quadratic polynomials.

(vii)The highest degree of x in 7x

3

is 3, so it is a cubic polynomials.

Video Explanation

7 0

2 years ago

> Question A normal distribution is observed from the number of points per game for a certain basketball player. If the mean

densk [106]

Answer:

The probability that in a randomly selected game, the player scored greater than 24 points is 0.0013 or 0.13%

Step-by-step explanation:

Given that

Mean = μ = 15 points

SD = σ = 3 points

For calculating probability for a data point, first of all we have to calculate the z-score of the value.

We have to find the probability of score greater than 24, then the z-score of 24 is:

z-score = (x-μ)/σ

z = (24-15)/3

z = 9/3

z = 3

Now we have to use the z-score table to find the probability of z<3 then it will be subtracted from 1 to find the probability of z>3

So,

$P(z3) = 1 - P(z$

Converting into percentage

0.0013 * 100 = 0.13%

Hence,

The probability that in a randomly selected game, the player scored greater than 24 points is 0.0013 or 0.13%

8 0

2 years ago

ABCD is a square with a perimeter of 56cm. Fill in the following blanks

miv72 [106K]

Answer:

2134

Step-by-step explanation:

6 0

2 years ago

A simple random sample of size n=250 individuals who are currently employed is asked if they work at home at least once per week

Levart [38]

Answer:

99% confidence interval for the population proportion of employed individuals is [0.104 , 0.224].

Step-by-step explanation:

We are given that a simple random sample of size n=250 individuals who are currently employed is asked if they work at home at least once per week.

Of the 250 employed individuals surveyed, 41 responded that they did work at home at least once per week.

Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of individuals who work at home at least once per week = $\frac{41}{250}$ = 0.164

n = sample of individuals surveyed = 250

Here for constructing 99% confidence interval we have used One-sample z proportion statistics.

So, 99% confidence interval for the population proportion, p is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at 0.5%

level of significance are -2.5758 & 2.5758}

P(-2.5758 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

99% confidence interval for p = [ $\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.164-2.5758 \times {\sqrt{\frac{0.164(1-0.164)}{250} } }$ , $0.164+2.5758 \times {\sqrt{\frac{0.164(1-0.164)}{250} } }$ ]

= [0.104 , 0.224]

Therefore, 99% confidence interval for the population proportion of employed individuals who work at home at least once per week is [0.104 , 0.224].

7 0

2 years ago