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2 years ago

A 1st degree polynomial function whose graph intercepts the y-axis at 8.

Mathematics

1 answer:

asambeis [7]2 years ago

4 0

Your answer can be anything in the form y = mx+8 where you replace m with any real number.

You start with y = mx+b, and then replace the b with the y intercept 8.

The y intercept is where the polynomial crosses the y axis.

The value of m does not matter. So you could have y = 2x+8 or y = 3x+8 for instance. Replace m with whatever your favorite number is.

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The ratio 5cm to 3m can be written in the form 1:n

arlik [135]

Answer:

Step-by-step explanation:

5/3=1/n(do cross multiplication)

3*1=5*n

3=5n

3/5=n

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2 years ago

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3 years ago

Find the value of x........................

Ann [662]

Answers: x = 4 and y = 3

============================================================

Explanation:

Triangle UBD is congruent to triangle RAN

Note the order of the lettering as it's important.

BD and AN are the last two letters of UBD and RAN in that order. Therefore, these sides correspond and BD = AN. From that, we know,

BD = AN

2y+5 = 3y+2

5-2 = 3y-2y

3 = 1y

y = 3

-----------

Using similar logic, UD and RN are the first and last letters of UBD and RAN respectively. So,

UD = RN

15 = 2x+7

2x+7 = 15

2x = 15-7

2x = 8

x = 8/2

x = 4

3 0

2 years ago

A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writin

Morgarella [4.7K]

Answer:

(a) We reject our null hypothesis.

(b) We fail to reject our null hypothesis.

(c) We fail to reject our null hypothesis.

Step-by-step explanation:

We are given that a certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writing machine) is at least 10 hr.

A random sample of 18 pens is selected.

Let $\mu$ = true average writing lifetime under controlled conditions

So, Null Hypothesis, $H_0$ : $\mu \geq$ 10 hr {means that the true average writing lifetime under controlled conditions is at least 10 hr}

Alternate Hypothesis, $H_A$ : $\mu$ < 10 hr {means that the true average writing lifetime under controlled conditions is less than 10 hr}

The test statistics that is used here is one-sample t test statistics;

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

where, $\bar X$ = sample mean

s = sample standard deviation

n = sample size of pens = 18

n - 1 = degree of freedom = 18 -1 = 17

Now, the decision rule based on the critical value of t is given by;

If the value of test statistics is more than the critical value of t at 17 degree of freedom for left-tailed test, then we will not reject our null hypothesis as it will not fall in the rejection region.

If the value of test statistics is less than the critical value of t at 17 degree of freedom for left-tailed test, then we will reject our null hypothesis as it will fall in the rejection region.

(a) Here, test statistics, t = -2.4 and level of significance is 0.05.

Now, at 0.05 significance level, the t table gives critical value of -1.74 at 17 degree of freedom.

Here, clearly the value of test statistics is less than the critical value of t as -2.4 < -1.74, so we reject our null hypothesis.

(b) Here, test statistics, t = -1.83 and level of significance is 0.01.

Now, at 0.051 significance level, the t table gives critical value of -2.567 at 17 degree of freedom.

Here, clearly the value of test statistics is more than the critical value of t as -2.567 < -1.83, so we fail to reject our null hypothesis.

(c) Here, test statistics, t = 0.57 and level of significance is not given so we assume it to be 0.05.

Now, at 0.05 significance level, the t table gives critical value of -1.74 at 17 degree of freedom.

Here, clearly the value of test statistics is more than the critical value of t as -1.74 < 0.57, so we fail to reject our null hypothesis.

8 0

2 years ago