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allochka39001 [22]
2 years ago
5

Masha is making a scale diagram of the front face of an aquarium. The face is shaped like a rectangle. She uses a scale of 1 cen

timeter to 6 inches to draw the diagram. The actual length of the aquarium face is 5 feet, and its width is 2 feet. The perimeter of the scale drawing of the front face is_______ centimeters.
Mathematics
1 answer:
dangina [55]2 years ago
4 0
14 feet because 5+5+2+2 equals 14
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2.161616 as fraction and as a mixed number
Hunter-Best [27]
2.161616 as a fraction is 2 161616/1000000 in simplest form its 2 10101/62500
3 0
2 years ago
Read 2 more answers
Three students, Linda, Tuan, and Javier, are given five laboratory rats each for a nutritional experiment. Each rat's weight is
Liula [17]

Answer:

The variance in weight is statistically the same among Javier's and Linda's rats

The null hypothesis will be accepted because the P-value (0.53 ) > ∝ ( level of significance )

Step-by-step explanation:

considering the null hypothesis that there is no difference between the weights of the rats, we will test the weight gain of the rats at 10% significance level with the use of Ti-83 calculator

The results from the One- way ANOVA  ( Numerator )

with the use of Ti-83 calculator

F = .66853

p = .53054

Factor

df = 2  ( degree of freedom )

SS = 23.212

MS = 11.606

Results from  One-way Anova ( denominator )

Ms = 11.606

Error

df = 12 ( degree of freedom )

SS = 208.324

MS = 17.3603

Sxp = 4.16657

where : test statistic = 0.6685

             p-value = 0.53

             level of significance ( ∝ ) = 0.10

The null hypothesis will be accepted because the P-value (0.53 ) > ∝

where Null hypothesis H0 = ∪1 = ∪2 = ∪3

hence The variance in weight is statistically the same among Javier's and Linda's rats

7 0
1 year ago
Roberto wants to display his 18 sports cards in an album. Some pages hold 2 cards and others hold 3 cards. How many different wa
WITCHER [35]
<span>65 = number of different arrangements of 2 and 3 card pages such that the total number of card slots equals 18. 416,154,290,872,320,000 = number of different ways of arranging 18 cards on the above 65 different arrangements of page sizes. ===== This is a rather badly worded question in that some assumptions aren't mentioned. The assumptions being: 1. The card's are not interchangeable. So number of possible permutations of the 18 cards is 18!. 2. That all of the pages must be filled. Since the least common multiple of 2 and 3 is 6, that means that 2 pages of 3 cards can only be interchanged with 3 pages of 2 cards. So with that said, we have the following configurations. 6x3 card pages. Only 1 possible configuration. 4x3 cards and 3x2 cards. These pages can be arranged in 7!/4!3! = 35 different ways. 2x3 cards and 6x2 cards. These pages can be arranged in 8!/2!6! = 28 ways 9x2 card pages. These can only be arranged in 1 way. So the total number of possible pages and the orders in which that they can be arranged is 1+35+28+1 = 65 possible combinations. Now for each of those 65 possible ways of placing 2 and 3 card pages such that the total number of card spaces is 18 has to be multiplied by the number of possible ways to arrange 18 cards which is 18! = 6402373705728000. So the total amount of arranging those cards is 6402373705728000 * 65 = 416,154,290,872,320,000</span>
6 0
2 years ago
A vehicle with a particular defect in its emission control system is taken to a succession of randomly selected mechanics until
IrinaK [193]

Answer:

the mle of P=0.833

Step-by-step explanation:

X=incorrect answer

And probability of success to be denoted as P

Here X posses a binomial distribution along with 'r' and 'p'parameter

CHECK THE ATTACHMENT BELOW FOR DETAILED EXPLATION

3 0
2 years ago
A region R in the xy-plane is given. Find equations for a transformation T that maps a rectangular region S in the uv-plane onto
Gre4nikov [31]
\begin{cases}y=2x-2\\y=2x+2\end{cases}\implies\begin{cases}-2x+y=-2\\-2x+y=2\end{cases}

For these lines, let u=-2x+y.

\begin{cases}y=2-x\\y=4-x\end{cases}\implies\begin{cases}x+y=2\\x+y=4\end{cases}

And for these, let v=x+y.

Now,

\begin{cases}u=-2x+y\\v=x+y\end{cases}\implies \begin{bmatrix}u\\v\end{bmatrix}=\underbrace{\begin{bmatrix}-2&1\\1&1\end{bmatrix}}_{\mathbf T}\begin{bmatrix}x\\y\end{bmatrix}

The vertices of S in the x-y plane are (0, 2), (2/3, 10/3), (2, 2), and (4/3, 2/3). Applying \mathbf T to each of these yields, respectively, (2, 2), (2, 4), (-2, 4), and (-2, 2), which are the vertices of a rectangle whose sides are parallel to the u-v plane.
6 0
2 years ago
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