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skad [1K]
2 years ago
14

A satellite makes 4 revolutions of the earth in one day. How many revolutions would it make in 6 1/2 days?

Mathematics
1 answer:
koban [17]2 years ago
8 0
26. 
1 day/24 hours, makes 4 revolutions. 
4 x 6 = 24, 
1/2 of 4 = 2.
24+2 = 26. 
26 revolutions, if my answer is wrong someone please correct me.
You might be interested in
Cory predicts it will take him 64 minutes to travel to Baltimore from Washington D.C. If the trip actually took 74 minutes, what
n200080 [17]

Answer:13.5



Step-by-step explanation:

the answer is 13.5


6 0
1 year ago
In 2009, a city with a population of 796,000 had 2,388 births. What is the number of births per 10,000 in the city?
Tom [10]
796,000/2388=10,000/x
796,000x= (2388*10,000)
796,000x/796,000=23,880,000/796,000
x=30 births
7 0
2 years ago
The local skating rink pays marry a fixed rate per pupil plus a base amount to work as a skating instructor. she earns $90 for i
Ganezh [65]
You have to make a system of equations: lets make a equal the amount marry makes per student and b be her base amount.
90=15a+b (you have to subtract the top equation by the bottom equation)
62=8a+b   (90-62=28, 15a-8a=7a, and b-b=0)
Since b canceled out, you are left with 7a=28 which means a=4.  you can than plug a into the equation 62=8a+b to find that b=30.

since Lisa makes half of the base amount marry, her base amount is 15.  However, she also make twice the amount per kid so she makes 8 per kid.
using the found values found you can make the equations (m=the amount Marry makes, l=the amount Lisa makes, and c is the number of children)
m=4c+30
l=8c+15
set c=20 and you should get m=110 and l=175.  Based off of that information, we can say that Lisa makes more money instructing a class of 20 students.
I hope this helps.
8 0
1 year ago
The equation A(t) = 900(0.85)t represents the value of a motor scooter t years after it was purchased. Which statements are also
ValentinkaMS [17]

Answer:

When new, the scooter cost $900

Step-by-step explanation:

<u><em>The complete question in the attached figure</em></u>

we have

A(t)=900(0.85)^t

This is a exponential function of the form

A(t)=a(b)^t

where

A(t) ----> represent the value of a motor scooter

t ----> the number of years after it was purchased

a ---> represent the initial value or y-intercept

b is the base of the exponential function

r is the percent rate of change

b=(1+r)

In this problem we have

a=\$900\\b=0.85

The base b is less than 1

That means ----> is a exponential decay function (is a decreasing function)

Find the percent rate of change

b=(1+r)\\0.85=1+r\\r=0.85-1\\r=-0.15

Convert to percentage (multiply by 100)

r=-15\% ---> negative means is a decreasing function

<u><em>Verify each statements</em></u>

<em>case A</em>) When new, the scooter cost $765.

The statement is false

Because the original value of the scooter was $900

case B) When new, the scooter cost $900

The statement is true (see the explanation)

case C) The scooter’s value is decreasing at a rate of 85%  each year

The statement is false

Because the scooter’s value is decreasing at a rate of 15%  each year (see the explanation)

case D) The scooter’s value is decreasing at a rate of  0.15% each year

The statement is false

Because the scooter’s value is decreasing at a rate of 15%  each year (see the explanation)

8 0
2 years ago
Three assembly lines are used to produce a certain component for an airliner. To examine the production rate, a random sample of
nikitadnepr [17]

Answer:

a) Reject H₀

b) [0.31; 3.35]

Step-by-step explanation:

Hello!

a) The objective of this example is to compare if the population means of the production rate of the assembly lines A, B and C. To do so the data of the production of each line were recorded and an ANOVA was run using it.

The study variable is:

Y: Production rate of an assembly line.

Assuming that the study variable has a normal distribution for each population, the observations are independent and the population variances are equal, you can apply a parametric ANOVA with the hypothesis:

H₀ μ₁= μ₂= μ₃

H₁: At least one of the population means is different from the others

Where:

Population 1: line A

Population 2: line B

Population 3: line C

α: 0.01

This test is always one-tailed to the right. The statistic is the Snedecor's F, constructed as the MSTr divided by the MSEr if the value of the statistic is big, this means that there is a greater variance due to the treatments than to the error, this means that the population means are different. If the value of F is small, it means that the differences between populations are not significant ( may differ due to error and not treatment).

The critical region is:

F_{k-1;n-k; 1-\alpha } = F_{2;15; 0.99} = 6.36

If F ≥ 3.36, the decision is to reject the null hypothesis.

Looking at the given data:

F= \frac{MSTr}{MSEr}= 11.32653

With this value the decision is to reject the null hypothesis.

Using the p-value method:

p-value: 0.001005

α: 0.01

The p-value is less than the significance level, the decision is to reject the null hypothesis.

At a level of 5%, there is significant evidence to say that at least one of the population means of the production ratio of the assembly lines A, B and C is different than the others.

b) In this item, you have to stop paying attention to the production ratio of the assembly line A to compare the population means of the production ratio of lines B and C.

(I'll use the same subscripts to be congruent with part a.)

The parameter to estimate is μ₂ - μ₃

The populations are the same as before, so you can still assume that the study variables have a normal distribution and their population variances are unknown but equal. The statistic to use under these conditions, since the sample sizes are 6 for both assembly lines, is a pooled-t for two independent variables with unknown but equal population variances.

t=  (X[bar]₂ - X[bar]₃) - ( μ₂ - μ₃) ~t_{n_2+n_3-2}

Sa√(1/n₂+1/n₃)

The formula for the interval is:

(X[bar]₂ - X[bar]₃) ± t_{n_2+n_3-2; 1 - \alpha /2}* Sa\sqrt{*\frac{1}{n_2} + \frac{1}{n_3} }

Sa^{2} = \frac{(n_2-1)*S_2^2+ (n_3-1)*S_3^2}{n_2+n_3-2}

Sa^{2} = \frac{(5*0.67)+ (5*0.7)}{6+6-2}

Sa^{2} = 0.685

Sa= 0.827 ≅ 0.83

t_{n_2+n_3-2;1-\alpha /2}= t_{10;0.995} = 3.169

X[bar]₂ = 43.33

X[bar]₃ = 41.5

(43.33-41.5) ± 3.169 * *0.83\sqrt{*\frac{1}{6} + \frac{1}{6} }

1.83 ± 3.169 * 0.479

[0.31; 3.35]

With a confidence level of 99% you'd expect that the difference of the population means of the production rate of the assemly lines B and C.

I hope it helps!

8 0
1 year ago
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