All categories

3 years ago

Y=1/2x + 3/4

1 answer:

maks197457 [2]3 years ago

4 0

to test the points, substitute the x-coordinate into the equation and if the value of y equals the y-coordinate then the point lies on the graph

(0, $\frac{1}{2}$ ) → y = 0 + $\frac{3}{4}$ = $\frac{3}{4}$ ≠ $\frac{1}{2}$

(0 , $\frac{3}{4}$ ) → y = $\frac{3}{4}$ ← on graph

( $\frac{3}{4}$ , 0) → y = $\frac{3}{8}$ + $\frac{3}{4}$ ≠ 0

( $\frac{3}{4}$ , $\frac{1}{2}$ → y = $\frac{3}{8}$ + $\frac{3}{4}$ ≠ $\frac{1}{2}$

( $\frac{1}{2}$ , 1) → y = $\frac{1}{4}$ + $\frac{3}{4}$ = 1 ← on graph

the points (0, $\frac{3}{4}$ ) and ( $\frac{1}{2}$ , 1) are on the graph

You might be interested in

Simplify |3 − 11| − (15 ÷ 3 + 2) 2

Vikentia [17]

|-8| - (5 + 2)2
8-(7)2
8-14= -6

Follow order of operations

6 0

3 years ago

PLEASE HELP ASAP! BRAINLIEST TO BEST/RIGHT ANSWER

Semenov [28]

= 9 * 10^5 * 3 * 10^10

= 27 * 10^5 * 10^10

= 27 * 10^15

= 2.7 * 10^16 - The best selection appears to be the first option.

Good luck :)

4 0

4 years ago

98 points possible

Pavel [41]

Answer:

Step-by-step explanation:

symmetry with respect to y-axis for y=f(x) means f(-x)=f(x)

in this case, y = f(x) = x / (x^2+4)

f(-x) = -x / ((-x)^2+4) = -x / (x^2+4) = -f(x)

so it is not symmetric to y-axis

symmetry with respect to x-axis for x=g(y) means g(-y)=g(y)

in this case, y = x / (x^2+4)

y*(x^2+4) = x

y*x^2 + 4y - x = 0

substitute -y into g(y)

(-y)*x^2 +4(-y) - x = 0

-y*x^2 - 4y - x = 0

y*x^2 + 4y + x = 0

so g(-y) is not equal to g(y)

so it is not symmetric to x-axis

8 0

3 years ago

Read 2 more answers

A sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.

TEA [102]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ = $1,150

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ $1,150

(b) The test statistic is 3.571.

(c) We conclude that the mean of all account balances is significantly different from $1,150.

Step-by-step explanation:

We are given that a sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.

We have test the hypothesis to determine whether the mean of all account balances is significantly different from $1,150.

Let $\mu$ = mean of all account balances

(a)So, Null Hypothesis, $H_0$ : $\mu$ = $1,150 {means that the mean of all account balances is equal to $1,150}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ $1,150 {means that the mean of all account balances is significantly different from $1,150}

The test statistics that will be used here is One-sample t test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average balance = $1,200

s = sample standard deviation = $126

n = sample of account balances = 81

(b) So, test statistics = $\frac{1,200-1,150}{\frac{126}{\sqrt{81} } }$ ~ $t_8_0$

= 3.571

The value of the sample test statistics is 3.571.

(c) Now, P-value of the test statistics is given by the following formula;

P-value = P( $t_8_0$ > 3.571) = Less than 0.05%

Because in the t table the highest critical value for t at 80 degree of freedom is given between 3.460 and 3.373 at 0.05% level.

Now, since P-value is less than the level of significance as 5% > 0.05%, so we sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the mean of all account balances is significantly different from $1,150.

8 0

4 years ago

If u know can u help me I got a lot of homework I just need answers for 11-15

Tju [1.3M]

For question 11
The median: 70
the mean: 66

3 0

3 years ago