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2 years ago

(SHOW UR WORK I DONT FEEL LIKE DOING THIS XDD BUT THANKSS) 98.73 ÷ 15 =

Mathematics

1 answer:

Juliette [100K]2 years ago

5 0

Answer:

Answer 6.582

Step-by-step explanation:

0 6. 5 8 2

1 5 9 8. 7 3 0

− 0

9 8

− 9 0

8 7

− 7 5

1 2 3

− 1 2 0

3 0

− 3 0

Hope its understandable ._.

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A student organization wondered how many hours per week college students study during a term. Each of the 10 students in the org

Inga [223]

Answer:

No, a convenience sample was taken.Therefore, the sample may not be representative of all college students in terms of number of hours studied per week.

Step-by-step explanation:

One of the important conditions required in other to estimate the mean of the population from the confidence interval. Requires that the data must be a random Sample. This is to ensure that sample values are not biased and as such are representative of the larger group of data.

However, in the scenario described above, the sample chose. Are friend's of the student, this makes the data lack randomness as those who aren't friends to the student do not stand a chance of being selected. The sampling procedure employed is called convenience sampling.

4 0

3 years ago

Please help me with this

sergij07 [2.7K]

Answer:

x=8

Step-by-step explanation:

The ratio of a side of PQRS to the corresponding side in TUVW. is 6:4=3:2.

The ratio of RS:VW=3:2. RS=12, and VW=x. 12:x=3:2. 3 multiplied by 4 is 12, so 2 multiplied by 4 is x. So, the makes x=8.

3 0

3 years ago

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

3 0

3 years ago

Joshua sells a pack of pens for $3.15, which is 5 percent more than he pays for them. Which equation will help find x, the amoun

vesna_86 [32]

Answer:

x + 0.05x = 3.15, which has one solution

Step-by-step explanation:

Let x = the price Joshua pays for the pack of pens

He add 5% (=.05 in decimal form)

That is x* .05

The new cost is 3.15

x+.05x = 3.15

This will have one solution

3 0

3 years ago

Read 2 more answers

Demarcus drew a pair of complementary angles he drew one angle that had a measure of 48* what was the measure of the other angle

Masja [62]

A complementary angle would be 90-x. in this case, x is 48°, so set up the equation 90-48 = other angle. this would give you 42, so the answer is A.

4 0

3 years ago