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3 years ago

HELP ME PLEASE,

Mathematics

2 answers:

jenyasd209 [6]3 years ago

6 0

Never is )4744$, $ :!83

bagirrra123 [75]3 years ago

6 0

I'm in K12 too, Those little things in the middle of the lesson do NOT need to be completed. You can skip over those. Also if you're on the grade level where you start having "discussions" those are not necessary either.

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Evaluate the line integral, where C is the given curve. (x + 6y) dx + x2 dy, C C consists of line segments from (0, 0) to (6, 1)

Dima020 [189]

Split $C$ into two component segments, $C_1$ and $C_2$ , parameterized by

$\mathbf r_1(t)=(1-t)(0,0)+t(6,1)=(6t,t)$

$\mathbf r_2(t)=(1-t)(6,1)+t(7,0)=(6+t,1-t)$

respectively, with $0\le t\le1$ , where $\mathbf r_i(t)=(x(t),y(t))$ .

We have

$\mathrm d\mathbf r_1=(6,1)\,\mathrm dt$

$\mathrm d\mathbf r_2=(1,-1)\,\mathrm dt$

where $\mathrm d\mathbf r_i=\left(\dfrac{\mathrm dx}{\mathrm dt},\dfrac{\mathrm dy}{\mathrm dt}\right)\,\mathrm dt$

so the line integral becomes

$\displaystyle\int_C(x+6y)\,\mathrm dx+x^2\,\mathrm dy=\left\{\int_{C_1}+\int_{C_2}\right\}(x+6y,x^2)\cdot(\mathrm dx,\mathrm dy)$

$=\displaystyle\int_0^1(6t+6t,(6t)^2)\cdot(6,1)\,\mathrm dt+\int_0^1((6+t)+6(1-t),(6+t)^2)\cdot(1,-1)\,\mathrm dt$

$=\displaystyle\int_0^1(35t^2+55t-24)\,\mathrm dt=\frac{91}6$

6 0

2 years ago

Convert the improper fraction 35/8 into a mixed number

kondor19780726 [428]

35/8 = (4 * 8 = 32) and 3 left over so answer is 4 3/8

7 0

3 years ago

Members of the millennial generation are continuing to be dependent on their parents (either living with or otherwise receiving

Morgarella [4.7K]

Answer:

$\bf H_0:$ The mean of adults aged 18 to 32 that continue to be dependent on their parents is 0.3

$\bf H_a:$ The mean of adults aged 18 to 32 that continue to be dependent on their parents is greater than 0.3

b) 34%

c) practically 0

d) Reject the null hypothesis.

Step-by-step explanation:

Since an individual aged 18 to 32 either continues to be dependent on their parents or not, this situation follows a Binomial Distribution and, according to the previous research, the probability p of “success” (depend on their parents) is 0.3 (30%) and the probability of failure q = 0.7

According to the sample, p seems to be 0.34 and q=0.66

To see if we can approximate this distribution with a Normal one, we must check that is not too skewed; this can be done by checking that np ≥ 5 and nq ≥ 5, where n is the sample size (400), which is evident.

We can then, approximate our Binomial with a Normal with mean

$\bf np = 400*0.34 = 136$

and standard deviation

$\bf \sqrt{npq}=\sqrt{400*0.34*0.66}=9.4742$

Since in the current research 136 out of 400 individuals (34%) showed to be continuing dependent on their parents:

$\bf H_0:$ The mean of adults aged 18 to 32 that continue to be dependent on their parents is 0.3

$\bf H_a:$ The mean of adults aged 18 to 32 that continue to be dependent on their parents is greater than 0.3

So, this is a right-tailed hypothesis testing.

According to the sample the proportion of "millennials" that are continuing to be dependent on their parents is 0.34 or 34%

Our level of significance is 0.05, so we are looking for a value $\bf Z^*$ such that the area under the Normal curve to the right of $\bf Z^*$ is ≤ 0.05

This value can be found by using a table or the computer and is $\bf Z^*$ = 1.645

Applying the continuity correction factor (this should be done because we are approximating a discrete distribution (Binomial) with a continuous one (Normal)), we simply add 0.5 to this value and

$\bf Z^*$ corrected is 2.145

Now we compute the z-score corresponding to the sample

$\bf z=\frac{\bar x -\mu}{s/\sqrt{n}}$