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denis-greek [22]
3 years ago
5

2 = 9 + 5/2 please answer this

Mathematics
2 answers:
Fantom [35]3 years ago
8 0

Answer:

false

Step-by-step explanation:

Reil [10]3 years ago
5 0

Answer:

Convert 529 5 2 9 to an improper fraction.

Step-by-step explanation:

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I’m confused please help
mart [117]

Answer:

im guessing the answer is 43.7

Step-by-step explanation:

7 0
3 years ago
Angles A = 60, B = 2x, C = 3x and D = 25 are 4 angles on a straight, what 3 points is the value of angle B?*​
iragen [17]

Answer:

  B = 38°

Step-by-step explanation:

A straight angle is 180°. If your four angles total a straight angle, then ...

  60 +2x +3x +25 = 180

  95 = 5x . . . . . . subtract 85

  x = 19 . . . . . . divide by 5

  2x = 38 . . . . . angle B is 38°

4 0
2 years ago
Epic math question that i cant solve
zvonat [6]
The area of the shaded region can be found by subtracting the area of the inside rectangle from the outside rectangle

area of outside: 6 X 10y —> 60y
area of inside: 3 X 5x —> 15x

so the area of the shaded region is 60y-15x

the GCF is 15, so the factored form is: 15(4y-x)

8 0
3 years ago
A study was conducted and two types of engines, A and B, were compared. Fifty experiments were performed using engine A and 75 u
USPshnik [31]

Answer:

a) -6 mpg.

b) 2.77 mpg

c) The 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results, in mpg, is (-8.77, -3.23).

Step-by-step explanation:

To solve this question, we need to understand the central limit theorem, and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Gas mileage A: Mean 36, standard deviation 6, sample of 50:

So

\mu_A = 36, s_A = \frac{6}{\sqrt{50}} = 0.8485

Gas mileage B: Mean 42, standard deviation 8, sample of 50:

So

\mu_B = 42, s_B = \frac{8}{\sqrt{50}} = 1.1314

Distribution of the difference:

Mean:

\mu = \mu_A - \mu_B = 36 - 42 = -6

Standard error:

s = \sqrt{s_A^2+s_B^2} = \sqrt{0.8485^2+1.1314^2} = 1.4142

A. Find the point estimate.

This is the difference of means, that is, -6 mpg.

B. Find the margin of error

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.025 = 0.975, so Z = 1.96.

Now, find the margin of error M as such

M = zs = 1.96*1.4142 = 2.77

The margin of error is of 2.77 mpg

C. Construct the 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results(5 pts)

The lower end of the interval is the sample mean subtracted by M. So it is -6 - 2.77 = -8.77 mpg

The upper end of the interval is the sample mean added to M. So it is -6 + 2.77 = -3.23 mpg

The 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results, in mpg, is (-8.77, -3.23).

8 0
3 years ago
1 question, Thanks if you answer, will mark brainliest.
ZanzabumX [31]
Ms Mess gave 2*100=200$
The Clerk gave 3*50=150$
200-150=50$
The end up paying 50-70= (-20$)
Not sure
6 0
3 years ago
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