Find the slope of the equation3x + 5y = 30

Please help me please please : ( : (

Gala2k [10]

Answer:

They will be picked up at 2 $\frac{1}{4}$ yards above sea level

Step-by-step explanation:

Let us consider sea level as reference and positions above sea level as positive and below sea level as negative.

With respect to this reference,

the position of top most point is +17 $\frac{3}{4}$ yards

and the position of lower most point is -13 $\frac{1}{4}$ yards

⇒ The position of midpoint is $\frac{+17 \frac{3}{4} - 13 \frac{1}{4}}{2}$

= +2 $\frac{1}{4}$

∴ <u>They will be picked up at 2 $\frac{1}{4}$ yards above sea level</u>

6 0

2 years ago

A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. (a) Describe the

Ivanshal [37]

Answer:

a) $\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

b) $z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

c) $z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

d) $z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

$\mu = 83, \sigma = 27$

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

$\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

Part b

We want this probability:

$P(\bar X>89)$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 89 we got:

$z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

Part c

$P(\bar X$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 75.65 we got:

$z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

Part d

We want this probability:

$P(79.4 < \bar X < 89.3)$

We find the z scores:

$z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

8 0

2 years ago

Which graph represents the function<br> f(x) = (x + 4)(x + 1)(x - 3)<br> ?

attashe74 [19]

Answer:

Step-by-step explanation:

The graph that represents the function

f(x) = (x + 4)(x + 1)(x - 3)

crosses the x-axis three times: at x = -4, x = -1, and x = 3

because we apply the zero product rule and found the roots -4, -1, and 3

x+ 4= 0 → x = -4

x+1 = 0 → x = -1

x-3 = 0 → x = 3

8 0

1 year ago

Urgent!!!!!!!plzz

Nitella [24]

Answer:

too much reading, can you put more points.

(edit: OML SO SORRY -THOUGHT THIS WAS QUESTIONS)

Step-by-step explanation:

7 0

2 years ago

The amount A of the radioactive element radium in a sample decays at a rate proportional to the amount of radium present. Given

slavikrds [6]

Answer:

a) $\frac{dm}{dt} = -k\cdot m$ , b) $m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$ , c) $m(t) = 10\cdot e^{-\frac{t}{2438.155} }$ , d) $m(300) \approx 8.842\,g$

Step-by-step explanation:

a) Let assume an initial mass m decaying at a constant rate k throughout time, the differential equation is:

$\frac{dm}{dt} = -k\cdot m$

b) The general solution is found after separating variables and integrating each sides:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $\tau$ is the time constant and $k = \frac{1}{\tau}$

c) The time constant is:

$\tau = \frac{1690\,yr}{\ln 2}$

$\tau = 2438.155\,yr$

The particular solution of the differential equation is:

$m(t) = 10\cdot e^{-\frac{t}{2438.155} }$

d) The amount of radium after 300 years is:

$m(300) \approx 8.842\,g$

4 0

2 years ago

Read 2 more answers