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3 years ago

Express in exponential form. logbN = x

2 answers:

7 0

log(b,N) = x converts to b^x = N. The base of the log is the base of the exponent. The exponent term is on its own side in the original log equation.

Answer: b^x = N

andrey2020 [161]3 years ago

3 0

Answer:

In exponential form given function is equal to $N=b^x$

Step-by-step explanation:

We are given with function,

$log_b\,^N=x$

We have to express it in exponential form.

Consider,

$log_b\,^N=x$

exponential both side to the power of b as base of given log function is b.

we get

$b^{log_b\,^N}=b^x$

$N=b^x$

Therefore In exponential form given function is equal to $N=b^x$

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URGENT! Which of the following correctly expresses sin(2θ)−sin(6θ) as a product?

777dan777 [17]

Answer:

−2sin(2θ)cos(4θ)

Step-by-step explanation:

sin(2∅)-sin(6∅)

Sum product Identity

sinα - sinβ = 2cos((α+β)/2)sin((α-β)/2)

Putting α= 2∅ and β= 6∅ in the identity we get

sin(2∅)-sin(6∅) = 2cos((2∅+6∅)/2)sin((2∅-6∅)/2)

=2cos(8∅/2)sin(-4∅/2)

=2cos(4∅)sin(-2∅)

= -2sin(2∅)cos(4∅)

7 0

4 years ago

Please help write number with answer next to it

Alika [10]

I don't understand what it is asking me to do

5 0

3 years ago

The weights of a certain brand of candies are normally distributed with a mean weight of 0.8556 g and a standard deviation of 0.

babunello [35]

Answer:

a) There is a probability P=0.4992 that a randomly selected candy weights at least 0.8543 g.

b) There is a probability P=0.3712 that a randomly selected sample of 441 candies have an average weight of at least 0.8543 g.

c) The claim of the brand should be re-evaluated as it is not providing consumers with the amount claimed on the label.

Step-by-step explanation:

The average weight of the candies in the package is:

$\bar x=\dfrac{\sum x_i}{n}=\dfrac{400.3}{469}=0.8535$

For a randomly selected candy (is a sample of size n=1) the standard deviation is the population standard deviation (sigma=0.0511 g).

The probabiltiy that is weights more than 0.8536 can be calculated with the z-score.

$z=\dfrac{X-\mu}{\sigma/\sqrt{n}}= \dfrac{0.8536-0.8535}{0.0511/\sqrt{1}}=\dfrac{0.0001}{0.0511}=0.002$

$P(X>0.8536)=P(z>0.002)=0.4992$

There is a probability P=0.4992 that a randomly selected candy weights at least 0.8536 g.

b. If the sample is now of n=441 candies, and we want to know the probability that the mean weight is at least 0.8543 g, the z-score needs to be recalculated:

$z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{0.8543-0.8535}{0.0511/\sqrt{441}}=\dfrac{0.0008}{0.0024}=0.3288$

$P(X_s>0.8543)=P(z>0.3288)=0.37115$

There is a probability P=0.3712 that a randomly selected sample of 441 candies have an average weight of at least 0.8543 g.

c. To be more confident about the claim that the mean weight is 0.8556 g, we can calculate the probability that, for a package of 469 candies and using the mean of 0.8535 g that we calculated before, the average weight is at least 0.8556 g.

$z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{0.8556-0.8535}{0.0511/\sqrt{469}}=\dfrac{0.0021}{0.0024}=0.89$