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Makovka662 [10]
3 years ago
15

I need help with these two questions please !

Mathematics
2 answers:
Lapatulllka [165]3 years ago
5 0
For the first question the answer is B. multiply 357.43 by 12 you'll get the answer $4,289.
son4ous [18]3 years ago
3 0

Answer:

the first own is C $392.20 no B

Step-by-step explanation:


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Is the following shape a right triangle? How do you know?
Marat540 [252]

Answer:

Yes

Step-by-step explanation:

A has a right angle because A on the triangle is turned at a 90 degree angle.

5 0
3 years ago
Read 2 more answers
In rhombus FGHJ if FG= 16 what's HJ?<br>​
ankoles [38]

tell me more? ^^

so  i thing is 13

#%

5 0
2 years ago
A bag contains 20 white marbles and 30 black marbles. if 7 marbles are chosen, what is the probability that there will be 2 whit
AlekseyPX
This is a hypergeometric distribution problem.
Population (N=50=W+B) is divided into two classes, W (W=20)  and B (B=30).
We calculate the probability of choosing w (w=2) white and b (b=5) black marbles.
Hypergeometric probability gives
P(W,B,w,b)=C(W,w)C(B,b)/(C(W+B,w+b)
where
C(n,r)=n!/(r!(n-r)!) the number of combinations of choosing r out of n objects.

Here
P(20,30,2,5)
=C(20,2)C(30,5)/(20+30, 2+5)
=190*142506/99884400
=0.2710

Alternatively, doing the combinatorics way:
#of ways to choose 2 from 20 =C(20,2)
#of ways to choose 5 from 30=C(30,5)
total #of ways = C(50,7)
P(20,30,2,5)=C(20,2)*C(30,5)/C(50,7)
=0.2710 
as before.
8 0
3 years ago
Factor into two binomials : 1–bx–x+b
earnstyle [38]

1-bx - x + b

x(1-b)- x(1-b)

( x-x) (1-b)

8 0
3 years ago
Read 2 more answers
Researchers collected data on the numbers of hospital admissions resulting from motor vehicle​ crashes, and results are given be
kenny6666 [7]

Answer:

This hypothesis test shows there is indeed a significant difference between the number of hospital admissions from motor vehicle crashes on Friday the 13th and the number of hospital admissions from motor vehicle crashes on Friday the 6th.

Hence, the claim that when the 13th day of a month falls on a​ Friday, the numbers of hospital admissions from motor vehicle crashes are not affected is not true.

Step-by-step explanation:

The missing data from the question

The numbers of hospital admissions from motor vehicle crashes

Friday the 6th || 10 | 8 | 4 | 4 | 2

Friday the 13th | 12 | 10 | 12 | 14 | 14

The null hypothesis would be that there is no significant evidence to conclude that the number of hospital admissions from motor vehicle crashes on Friday the 6th is different from the number of hospital admissions from motor vehicle crashes on Friday the 13th. That is, when the 13th day of a month falls on a​ Friday, the numbers of hospital admissions from motor vehicle crashes are not affected.

The alternative hypothesis is that there is significant evidence to conclude that the number of hospital admissions from motor vehicle crashes on Friday the 6th is different from the number of hospital admissions from motor vehicle crashes on Friday the 13th. That is, when the 13th day of a month falls on a​ Friday, the numbers of hospital admissions from motor vehicle crashes are affected.

If the population mean of number of hospital admissions from motor vehicle crashes on Friday the 13th = μ₁

And the population mean of number of hospital admissions from motor vehicle crashes on Friday the 6th = μ₂

If the difference in population mean of number of hospital admissions from motor vehicle crashes on Friday the 13th and number of hospital admissions from motor vehicle crashes on Friday the 6th is μ

μ = μ₁ - μ₂

Mathematically,

The null hypothesis is that

H₀: μ = 0

Or

H₀: μ₁ = μ₂

The alternative hypothesis is given as

Hₐ: μ ≠ 0

Or

Hₐ: μ₁ ≠ μ₂

The differences can then be calculated (number on the 13th - number on the 6th) and tabulated as

Friday the 6th || 10 | 8 | 4 | 4 | 2

Friday the 13th | 12 | 10 | 12 | 14 | 14

Differences ||| 2 | 2 | 8 | 10 | 12

We need the sample mean and sample standard deviation.

Mean = (Σx)/N

= (2+2+8+10+12)/5 = 6.80

Standard deviation = σ = √[Σ(x - xbar)²/N]

Σ(x - xbar)² = (2-6.8)² + (2-6.8)² + (8-6.8)² + (10-6.8)² + (12-6.8)² = 84.8

σ = √[Σ(x - xbar)²/N] = √(84.8/5) = 4.12

To do this test, we will use the t-distribution because no information on the population standard deviation is known

So, we compute the t-test statistic

t = (x - μ)/σₓ

x = sample mean of difference = 6.80

μ = 0 (should be 0 if there's no difference between the two sets of data)

σₓ = standard error = [σ√n]

where n = Sample size = 5

σ = 4.12

σₓ = [4.12/√5] = 1.84

t = (6.80 - 0) ÷ 1.84

t = 3.70

checking the tables for the p-value of this t-statistic

Degree of freedom = df = n - 1 = 5 - 1 = 4

Significance level = 0.05

The hypothesis test uses a two-tailed condition because we're testing in two directions.

p-value (for t = 3.70, at 0.05 significance level, df = 4, with a two tailed condition) = 0.020835

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.05

p-value = 0.020835

0.020835 < 0.05

Hence,

p-value < significance level

This means that reject the null hypothesis, accept the alternative hypothesis & say that there is enough evidence to conclude that number of hospital admissions from motor vehicle crashes on Friday the 6th is different from the number of hospital admissions from motor vehicle crashes on Friday the 13th. That is, when the 13th day of a month falls on a​ Friday, the numbers of hospital admissions from motor vehicle crashes are affected.

Hope this Helps!!!

6 0
3 years ago
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