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grigory [225]
3 years ago
11

WILL GIVE A BRAINLIEST IF UR RIGHT!!!

Mathematics
1 answer:
Yuri [45]3 years ago
4 0
Power rule I believe.

w²w³ = w²⁺³ = w⁵
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Rewrite the following numbers in their best form
timurjin [86]
A. 70.105
b. 950
c. 0.0072
5 0
4 years ago
Tumor counts: A cancer laboratory is estimating the rate of tumorigenesis in two strains of mice, A and B. They have tumor count
Igoryamba

Answer:

The observed tumor counts for the two populations of mice are:

Type A mice = 10 * 12 = 120 counts

Type B mice = 13 * 12 = 156 counts

Step-by-step explanation:

Since type B mice are related to type A mice and given that type A mice have tumor counts that are approximately Poisson-distributed with a mean of 12, we can then assume that the mean of type A mice tumor count rate is equal to the mean of type B mice tumor count rate.

This is because the Poisson distribution can be used to approximate the the mean and variance of unknown data (type B mice count rate) using known data (type A mice tumor count rate).  And the Poisson distribution gives the probability of an occurrence within a specified time interval.

3 0
3 years ago
If sin theta = 8/17 and cot theta < 0, what is sec theta?
klasskru [66]

Answer:

-\frac{17}{15}

Step-by-step explanation:

By definition, \cot \theta=\frac{1}{\tan \theta} and \sec \theta=\frac{1}{\cos \theta}. Since since \cot \theta is negative, \tan \theta must also be negative, and since \sin \theta is positive, we must be in Quadrant II.

In a right triangle, the sine of an angle is equal to its opposite side divided by the hypotenuse. The cosine of an angle in a right triangle is equal to its adjacent side divided by the hypotenuse. Therefore, we can draw a right triangle in Quadrant II, where the opposite side to angle theta is 8 and the hypotenuse of the triangle is 17.

To find the remaining leg, use to the Pythagorean Theorem, where a^2+b^2=c^2, where c is the hypotenuse, or longest side, of the right triangle and a and b are the two legs of the right triangle.

Solving, we get:

8^2+b^2=17^2,\\b^2=17^2-8^2,\\b^2=\sqrt{17^2-8^2}=\sqrt{225}=15

Since all values of cosine theta are negative in Quadrant II, all values of secant theta must also be negative in Quadrant II.

Thus, we have:

\sec\theta=\frac{1}{\cos \theta}=-\frac{1}{\frac{15}{17}}=\boxed{-\frac{17}{15}}

6 0
3 years ago
I need help on number 2​
musickatia [10]
Start from -2,0 and then go up 1 and 4 to the right for each point
4 0
3 years ago
How to solve for the sin 11pi/2 without using a calculator?
vesna_86 [32]
You have to do some trigonometric transformations :
sin ( 11π/2 ) = sin ( 8π/2 + 3π/2 ) = sin ( 4π + 3π / 2 ) = sin ( 3π / 2 ) =
= sin 270° = -1
6 0
3 years ago
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