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hoa [83]
3 years ago
10

Someone smart help me please

Mathematics
1 answer:
mafiozo [28]3 years ago
7 0
Pretty sure it is
shift 1 left
i might be wrong
You might be interested in
Which is an x-intercept of the continuous function in the table?
Ad libitum [116K]

Answer:

B. (3, 0)

Step-by-step explanation:

The x-intercept is the point where the graph of the function meets the x-axis.

At x-intercept, y=0 or f(x)=0

So look through the table and find where f(x)=0.

From the table, f(x)=0 at x=3.

We write this as an ordered pair.

Therefore the x-intercept is (3,0)

The correct choice is B.

7 0
3 years ago
Read 2 more answers
I need help on this one please
TiliK225 [7]

Hi

Income tax is  progressive

So with  50 000  we have  :  

0.02*3000 + 0.03 ( 5000 -3001) + 0.05 ( 17 000-5001) + 0.0575 ( 50 000 - 17 001) =  60+ 59.97+599.95 + 1897.44 =  2617.36

tax is = 2617.36

5 0
3 years ago
Given the probability distributions shown to the​ right, complete the following parts.
Elan Coil [88]

Answer:

a) Expected Value for distribution A, E(X) = 3.020

Expected Value for distribution B, E(X) = 0.980

b) Standard deviation of distribution A = 1.157

Standard deviation of distribution B = 1.157

c) In distribution A, the bigger values of x have a higher probability of occurring than the values of distribution B (whose smaller values of x have a higher chance of occurring, hence, the expected value for distribution A is more than that of distribution B.

But according to the corresponding distribution of values, the two distributions have the same exact spread, A in ascending order (with higher values with bigger probability) and B in descending order (lower values have higher probabilities). But the same spread regardless, hence, the standard deviation which shows how data values spread around the mean (centre point) of a distribution is the same for the two distributions.

Step-by-step explanation:

Expected values is given as

E(X) = Σ xᵢpᵢ

where xᵢ = each possible sample space

pᵢ = P(X=xᵢ) = probability of each sample space occurring.

Distributions A and B is given by

X P(X) X P(x)

0 0.04 0 0.47

1 0.09 1 0.25

2 0.15 2 0.15

3 0.25 3 0.09

4 0.47 4 0.04

For distribution A

E(X) = Σ xᵢpᵢ = (0×0.04) + (1×0.09) + (2×0.15) + (3×0.25) + (4×0.47) = 3.02

For distribution B

E(X) = Σ xᵢpᵢ = (0×0.47) + (1×0.25) + (2×0.15) + (3×0.09) + (4×0.04) = 0.98

b) Standard deviation = √(variance)

But Variance is given by

Variance = Var(X) = Σx²p − μ²

where μ = E(X)

For distribution A

Σx²p = (0²×0.04) + (1²×0.09) + (2²×0.15) + (3²×0.25) + (4²×0.47) = 10.46

Variance = Var(X) = 10.46 - 3.02² = 1.3396

Standard deviation = √(1.3396) = 1.157

For distribution B

Σx²p = (0²×0.47) + (1²×0.25) + (2²×0.15) + (3²×0.09) + (4²×0.04) = 2.30

Variance = Var(X) = 2.30 - 0.98² = 1.3396

Standard deviation = √(1.3396) = 1.157

c) In distribution A, the bigger values of x have a higher probability of occurring than the values of distribution B (whose smaller values of x have a higher chance of occurring, hence, the expected value for distribution A is more than that of distribution B.

But according to the corresponding distribution of values, the two distributions have the same exact spread, A in ascending order (with higher values with bigger probability) and B in descending order (lower values have higher probabilities). But the same spread regardless, hence, the standard deviation which shows how data values spread around the mean (centre point) of a distribution is the same for the two distributions.

7 0
3 years ago
Write an equation in slope-intercept form of a line that passes through the point (8,-0) and is perpendicular to the line x - 3y
IceJOKER [234]

Step-by-step explanation:

let eqn be y = mx + b.

3y = x - 15

y = 1/3x - 5

perpendicular means product of grad = -1

hence m = -3.

sun (8, 0):

8 = -3(0) + b

b = 8

therefore, equation if the line is y = -3x + 8

Topic: coordinate geometry

If you like to venture further, feel free to check out my insta (learntionary). I'll be constantly posting math tips and notes! Thanks!

7 0
3 years ago
Is the mean weight of female college students still 59.4 kg? To test this, you take a random sample of 20 students, finding a me
adell [148]

Answer:

Correct option: (B)

Step-by-step explanation:

A one-sample <em>t</em>-test can be performed to determine whether the mean weight of female college students is still 59.4 kg.

The hypothesis can be defined as:

<em>H₀</em>: The mean weight of female college students has changed.

<em>Hₐ</em>: The mean weight of female college students has not changed.

The information provided is:

\bar x=61.65\\s=6.39\\n=20\\\alpha =0.10

As the there is no information about the population standard deviation we will use a <em>t</em>-test for the mean.

The test statistic is:

t=\frac{\bar x-\mu}{\s/\sqrt{n}}=\frac{61.65-59.4}{6.39/\sqrt{20}}=1.57

Decision rule:

If the if the <em>p</em>-value of the test is less than the significance level of the test <em>α</em> then the null hypothesis will be rejected and vice versa.

The degrees of freedom of the test is:

df=n-1=20-1=19

The test is two-tailed.

Compute the <em>p</em>-value of the test as follows:

p-value=2\times P(t_{0.10/2, 19}

*Use a <em>t</em>-table for the probability.

The <em>p</em>-value = 0.13 > <em>α</em> = 0.10

The null hypothesis was failed to be rejected.

As the null hypothesis was rejected it can be concluded that there is not sufficient evidence that the mean weight of female students has changed.

6 0
4 years ago
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