Write an equation in slope intercept form that passes through (2,1) and is perpendicular to (5x+y=9)

2 Quick algebra 1 question for 50 points!

SVETLANKA909090 [29]

#6

Yes

Direct variation because Time increases cupcakes increases and vice versa

#B

Constant of proportionality represents the amount of time required per cupcakes

#7

Yes we can find it

We have to check the x and y values of the ordered pairs (x,y)

If y is decreasing with respect to increase in x then it's inverse variation

3 0

2 years ago

The height of a free falling object at time t can be found using the function, h(t) = - 12t2 + 36t. Where h(t) is the height in

never [62]

For this case we have the following equation:
h (t) = - 12t2 + 36t
When the object hits the ground we have:
- 12t2 + 36t = 0
We look for the roots of the polynomial:
t1 = 0
t2 = 3
Therefore, the time it takes the object to hit the ground is:
t = 3 s
Answer:
the time when the object hits the ground is:
t = 3 s

7 0

3 years ago

Read 2 more answers

A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime

MrRissso [65]

Answer:

We conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

We conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

Step-by-step explanation:

We are given a random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen;

1.10, 5.09, 0.97, 1.59, 4.60, 0.32, 0.55, 1.45, 0.14, 4.47, 1.20, 3.50, 5.02, 4.67, 5.22, 2.69, 3.98, 3.17, 3.03, 2.21, 0.69, 4.47, 3.31, 1.17, 0.76, 1.17, 1.57, 2.62, 1.66, 2.05.

Let $\mu$ = true average percentage of organic matter

So, Null Hypothesis, $H_0$ : $\mu$ = 3% {means that the true average percentage of organic matter in such soil is 3%}

Alternate Hypothesis, $H_A$ : $\mu \neq$ 3% {means that the true average percentage of organic matter in such soil is something other than 3%}

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

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

where, $\bar X$ = sample mean percentage of organic matter = 2.481%

s = sample standard deviation = 1.616%

n = sample of soil specimens = 30

So, the test statistics = $\frac{2.481-3}{\frac{1.616}{\sqrt{30} } }$ ~ $t_2_9$

= -1.76

The value of t-test statistics is -1.76.

(a) Now, at 10% level of significance the t table gives a critical value of -1.699 and 1.699 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics doesn't lie within the range of critical values of t, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

(b) Now, at 5% level of significance the t table gives a critical value of -2.045 and 2.045 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

8 0

3 years ago

Ralph is an electrician. He charges an initial fee of $26, plus $31 per hour. If Ralph earned $119 on a job, how long did the jo

lubasha [3.4K]

3 hours

You can turn this into an equation:

F = 26 + 31t
Where F is the money Ralph earned, and t is the number of hours he worked.

Plugging in the 119 for F, we get
119 = 26 + 31t

93 = 31t

3 = t

5 0