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3 years ago

Simplify x2y5 ____ x6y3

Mathematics

1 answer:

viva [34]3 years ago

6 0

Answer:

y^2/x^4

i definitely recommend practicing this bc it's on the final exam for sure

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The top of a ladder rests at a height of 15 feet against the side of a house.If the base of the ladder is 6 feet from the house,

Lemur [1.5K]

21 feet because yea yessir

8 0

3 years ago

NEED ASAP MATH!! 7TH GRADE

lozanna [386]

Answer:

Step-by-step explanation:

4 0

3 years ago

What does it mean if the distance between a point p and a line l is zero? What does it mean if the distance between two lines is

-Dominant- [34]

Answer:

a) the point is on the line

b) the lines are the same line

Step-by-step explanation:

a) The distance from a point to a line is zero when the point satisfies the equation of the line. That is, the point is on the line.

b) Two lines are zero distance apart when they are on top of each other, indistinguishable. That is, they are the same line.

4 0

3 years ago

In a random sample of 400 residents of Boston, 320 residents indicated that they voted for Obama in the last presidential electi

coldgirl [10]

Answer:

C.I = 0.7608 ≤ p ≤ 0.8392

Step-by-step explanation:

Given that:

Let consider a random sample n = 400 candidates where 320 residents indicated that they voted for Obama

probability $\hat p = \dfrac{320}{400}$

= 0.8

Level of significance ∝ = 100 -95%

= 5%

= 0.05

The objective is to develop a 95% confidence interval estimate for the proportion of all Boston residents who voted for Obama.

The confidence internal can be computed as:

$=\hat p \pm Z_{\alpha/2} \sqrt{\dfrac{ p(1-p)}{n } }$

where;

$Z_{0.05/2}$ = $Z_{0.025}$ = 1.960

SO;

$=0.8 \pm 1.960 \sqrt{\dfrac{ 0.8(1-0.8)}{400 } }$

$=0.8 \pm 1.960 \sqrt{\dfrac{ 0.8(0.2)}{400 } }$

$=0.8 \pm 1.960 \sqrt{\dfrac{ 0.16}{400 } }$

$=0.8 \pm 1.960 \sqrt{4 \times 10^{-4}}$

$=0.8 \pm 1.960 \times 0.02}$

$=0.8 \pm 0.0392$

= 0.8 - 0.0392 OR 0.8 + 0.0392

= 0.7608 OR 0.8392

Thus; C.I = 0.7608 ≤ p ≤ 0.8392

3 0

3 years ago

Tank 1 initially contains 50 gals of water with 10 oz of salt in it, while tank 2 initially contains 20 gals of water with 15 oz

Naddik [55]

$\dfrac{\mathrm dx_1}{\mathrm dt}=\dfrac{2\text{ oz}}{1\text{ gal}}\dfrac{5\text{ gal}}{1\text{ min}}-\dfrac{x_1(t)\text{ oz}}{50\text{ gal}}\dfrac{5\text{ gal}}{1\text{ min}}$
$\dfrac{\mathrm dx_2}{\mathrm dt}=\dfrac{x_1(t)\text{ oz}}{50\text{ gal}}\dfrac{5\text{ gal}}{1\text{ min}}-\dfrac{x_2(t)\text{ oz}}{20\text{ gal}}\dfrac{5\text{ gal}}{1\text{ min}}$

$\implies\begin{cases}\dfrac{\mathrm dx_1}{\mathrm dt}=10-\dfrac1{10}x_1\\\\\dfrac{\mathrm dx_2}{\mathrm dt}=\dfrac1{10}x_1-\dfrac14x_2\\\\x_1(0)=10\\\\x_2(0)=15\end{cases}$

6 0

3 years ago