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3 years ago

Solve. -3 divided by 1/6

Mathematics

2 answers:

Ivan3 years ago

6 0

Answer:

-18

Step-by-step explanation:

suter [353]3 years ago

6 0

Answer:

-18

Step-by-step explanation:

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you can afford a $250 per month car payment. you've found a 3 year loan at 2% interest . how big of a loan can you afford

Elina [12.6K]

Answer:

interest = 250

time = 3 years

rate = 2%

we know

P = I*100/(R*T)

= 250*100/6

=4166.67

Step-by-step explanation:

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4 years ago

Samuel scored a 72, 92 and 80 on his first three exams. What is the minimum score he needs to get on his exam to get an average

sattari [20]

Samuel would need to score a minimum of 96 on his 4th exam

4 0

1 year ago

In a class of 42 students the number of boys is 2/5 of the girls . find the number of boys and girls

Ksju [112]

In plain and short, we simply divide 42 by (2+5) and distribute accordingly.

$\bf \cfrac{boys}{girls}\qquad 2:5\qquad \cfrac{2\cdot \frac{42}{2+5}}{5\cdot \frac{42}{2+5}}\implies \cfrac{2\cdot 6}{5\cdot 6}\implies \cfrac{12}{30}$

3 0

3 years ago

F(x, y, z) = z tan−1(y2)i + z3 ln(x2 + 3)j + zk. find the flux of f across s, the part of the paraboloid x2 + y2 + z = 18 that l

Cerrena [4.2K]

$\mathbf F(x,y,z)=z\tan^{-1}(y^2)\,\mathbf i+z^3\ln(x^2+3)\,\mathbf j+z\,\mathbf k$
$\implies\mathrm{div}\mathbf F(x,y,z)=0+0+1=1$

By the divergence theorem, the flux of $\mathbf F$ across the *closed* surface $\mathcal S$ combined with the plane $z=2$ is given by a volume integral over the closed region:

$\displaystyle\iint_{\mathcal S}\mathbf F\cdot\mathrm d\mathbf S=\iiint_{\mathcal R}\nabla\cdot\mathbf F\,\mathrm dV$

So in fact, to find the flux over $\mathcal S$ alone, we'll need to subtract the flux of $\mathbf F$ over the planar portion, oriented outward. First, compute the volume integral by converting to cylindrical coordinates:

$x^2+y^2+z=18$
$z=2\implies x^2+y^2=16\implies r^2=16\implies r=4$

$\displaystyle\iiint_{\mathcal R}\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=4}\int_{z=2}^{z=18-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=128\pi$

If the surface does actually contain $z=2$ , then you can stop here; otherwise, continue.

Now, parameterize the part of the *closed* surface in $z=2$ by

$\mathbf s(r,\theta)=r\cos\theta\,\mathbf i+r\sin\theta\,\mathbf j+2\,\mathbf k$

where $0\le r\le4$ and $0\le\theta\le2\pi$ . We get a surface element

$\mathrm d\mathbf S=(\mathbf s_r\times\mathbf s_\theta)\,\mathrm dr\,\mathrm d\theta=(r\,\mathbf k)\,\mathrm dr\,\mathrm d\theta$

We don't need to worry about the first two components of

and so the surface integral over this region is

$\displaystyle\iint_{z=2\,\land\,x^2+y^2\le16}\mathbf F\cdot\mathrm d\mathbf S=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=4}2r\,\mathrm dr\,\mathrm d\theta=32\pi$

Then the total flux over $\mathcal S$ alone is $(128-32)\pi=96\pi$ .

4 0

3 years ago

Suppose 462 of 500 randomly selected college students said they would be embarrassed to truthfully admit they could not read at

ANEK [815]

Answer:

We conclude that there is a significant difference in the proportion of all college students who would be embarrassed by these two admissions.

Step-by-step explanation:

We are given that 462 of 500 randomly selected college students said they would be embarrassed to truthfully admit they could not read at a fourth grade level.

Further suppose 135 of 500 randomly selected college students said they would be embarrassed to truthfully admit they could not "do math" at a fourth grade level (like fractions).

Let $p_1$ = proportion of college students who would be embarrassed to truthfully admit they could not read at a fourth grade level.

$p_2$ = proportion of college students who would be embarrassed to truthfully admit they could not "do math" at a fourth grade level.

So, Null Hypothesis, $H_0$ : $p_1-p_2$ = 0 or $p_1= p_2$ {means that there is not any significant difference in the proportion of all college students who would be embarrassed by these two admissions}

Alternate Hypothesis, $H_A$ : $p_1-p_2$ $\neq$ 0 or $p_1\neq p_2$ {means that there is a significant difference in the proportion of all college students who would be embarrassed by these two admissions}

The test statistics that would be used here Two-sample z proportion statistics;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of college students who would be embarrassed to admit they could not read at a fourth grade level = $\frac{462}{500}$ = 0.924

$\hat p_2$ = sample proportion of college students who would be embarrassed to admit they could not "do math" at a fourth grade level = $\frac{135}{500}$ = 0.27

$n_1$ = sample of college students = 500

$n_2$ = sample of college students = 500

So, test statistics = $\frac{(0.924-0.27)-(0)}{\sqrt{\frac{0.924(1-0.924)}{500}+\frac{0.27(1-0.27)}{500} } }$

= 28.28

The value of z test statistics is 28.28.

Also, P-value of the test statistics is given by;

P-value = P(Z > 28.28) = Less than 0.0005%

Now, at 0.10 significance level the z table gives critical values of -1.645 and 1.645 for two-tailed test.

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

Therefore, we conclude that there is a significant difference in the proportion of all college students who would be embarrassed by these two admissions.

5 0

4 years ago