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Vedmedyk [2.9K]

2 years ago

8

A Cepheid star is a type of variable star, which means its brightness is not constant. The relationship between the brightness o

f a Cepheid star and its period, or length of its pulse, is given by M = –2.78(log P) – 1.35, where M is the absolute magnitude, or brightness, of the star, and P is the number of days required for the star to complete one cycle. What is the absolute magnitude of a star that has a period of 45 days? Use a calculator. Round your answer to the nearest hundredth. –5.95 –4.60 3.25 4.60

2 answers:

White raven [17]2 years ago

8 0

Answer:

The absolute brightness of the Cepheid star after a period of 45 days is -5.95

Step-by-step explanation:

Since the absolute magnitude or brightness of a Cepheid star is related to its period or length of its pulse by

M = –2.78(log P) – 1.35 where M = absolute magnitude and P = period or length of pulse.

From our question, it is given that P = 45 days.

So, M = –2.78(log P) – 1.35

M = –2.78(log 45) – 1.35

M = –2.78(1.6532) – 1.35

M = -4.60 - 1.35

M = -5.95

So, the absolute magnitude or brightness M of a Cepheid star after a period P of 45 days is -5.95

BaLLatris [955]2 years ago

7 0

Answer:

A

Step-by-step explanation:

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A little girl is watching planes take off of the runway from a building's rooftop 40 meters away from the airport. If the height

worty [1.4K]

Answer:

h = 417.81 meters

The altitude of the plane is 417.81 meters

Step-by-step explanation:

given;

The initial height of the girl h0 = 400 m

Horizontal distance from the airport d = 40 m

Angle of elevation of the plane ∅= 24°

The altitude of the plane when the picture was taken is equal to initial height of the girl plus the elevation height.

h = he + h0

And he (elevation height) can be derived as;

he = dtan∅

he = 40tan24°

Substituting into the equation;

h = 40tan24° + 400

h = 417.81 meters

The altitude of the plane is 417.81 meters

5 0

2 years ago

A sales completion team, aiming to reduce the shipment time of urgent orders, studies the current process and finds that the cur

4vir4ik [10]

Answer:

$P(1$

And we can find this probability with thi difference:

$P(-2$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

$P(-2$

So then we expect about 72.5% of the current shipment times between 1 and 5 days

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the current shipment time of a population, and for this case we know the distribution for X is given by:

$X \sim N(4,1.5)$

Where $\mu=4$ and $\sigma=1.5$

We are interested on this probability

$P(1$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(1$

And we can find this probability with thi difference:

$P(-2$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

$P(-2$

So then we expect about 72.5% of the current shipment times between 1 and 5 days

8 0

2 years ago

Which statements are true regarding triangle LMN? Check all that apply.

dimaraw [331]

Answer:

$NM = x$

$LM = x\sqrt{2}$

$tan (45) = 1$

Step-by-step explanation:

Step 1: Pythagoras Theorem

Pythagoras theorem relates the three sides of the triangle in such a way that the sum of the square of base and perpendicular is equal to hypotenuse, such as:

$LM^{2} =LN^{2} +NM^{2}$

Step 2: Trigonometric Functions

Only for a right angle triangle following three trigonometric relations are valid

$sin (\theta) = \frac{opposite}{hypotenuse}$

$cos (\theta) = \frac{adjacent}{hypotenuse}$

$tan (\theta)=\frac{sin (\theta)}{cos (\theta)} = \frac{opposite}{adjacent}$

Step 3: Verifying all the possible answers

A: Since, LN = x and using $tan (45) =1$

we can calculate

$tan (\theta)= \frac{opposite}{adjacent}$

$tan (45)= \frac{NM}{x} =1$

therefore, NM = x (true)

B: As NM = x therefore it can not be equal to $x\sqrt{2\\}$ .

C: Using Pythagoras Theorem

$LM^{2} =LN^{2} +NM^{2}$

$LM^{2} =x^{2} +x^{2}$

$LM^{2} =2x^{2}$

$LM = \sqrt{2x^{2}} = x\sqrt{2}$

It can also be proved using trigonometric relation

$cos (45) = \frac{x}{LM}$

$LM = \frac{x}{cos (45)}$

As, $\frac{1}{cos (45)}= \sqrt{2}$

Therefore

$LM = x\sqrt{2}$

D and E:

Using same approach similar to part A

Since, LN = x and NM = x

we can calculate

$tan (\theta)= \frac{opposite}{adjacent}$

$tan (45)= \frac{x}{x} =1$

Therefore, $tan (45) = 1$ and not equal to $\frac{\sqrt{2} }{2}$

3 0

2 years ago

Read 2 more answers

The dimensions of a storage container in the shape of a rectangular prism are 54 in x 32.25 in x 24.5 in. Choose True or False f

AlladinOne [14]

GIVEN

the dimensions ofrectangular prism are

length = 54 in

breadth = 32.25in

height = 24.5 in

formula

volume of the rectangular prism = length×breadth×height

= 54 ×32.25×24.5

= 42997.5 in³

first statement is false because the volume of the rectangular prism in two significant digit.

Second statement is false because the volume of the rectangular prism in two significant digit.

FORMULA

surface area of a rectangular prism = 2(wl+hl+hw)

by putting the value of the length , breadth & height

we get

= 2 (54×32.25 + 32.25×24.5 + 24.5×54)

= 7709.25 in²

= 7709.3 in²(approx)

Third statement is true the surface area of the container is rounded to ones place.

Hence proved

3 0

2 years ago

Read 2 more answers

A team averaging 110 points is likely to do very well during the regular season. The coach of your team has hypothesized that yo

aev [14]

Answer:

1. M=108

2. μ=110

3. In the explanation.

4. Test statistic t = -1.05

5. P-value = 0.1597

Step-by-step explanation:

The question is incomplete: to solve this problem, we need the sample information: size, mean and standard deviation.

We will assume a sample size of 10 matches, a sample mean of 108 points and a sample standard deviation of 6 points.

1. The mean points is the sample points and has a value of 108 points.

2. The null hypothesis is H0: μ=110, meaning that the mean score is not significantly less from 110 points.

3. This is a hypothesis test for the population mean.

The claim is that the mean score is significantly less than 110.

Then, the null and alternative hypothesis are:

$H_0: \mu=110\\\\H_a:\mu< 110$

The significance level is 0.05.

The sample has a size n=10.

The sample mean is M=108.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=6.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{6}{\sqrt{10}}=1.9$

4. Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{108-110}{1.9}=\dfrac{-2}{1.9}=-1.05$

The degrees of freedom for this sample size are:

$df=n-1=10-1=9$

5. This test is a left-tailed test, with 9 degrees of freedom and t=-1.05, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t$

As the P-value (0.1597) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the mean score is significantly less than 110.

7 0

2 years ago