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AlekseyPX
3 years ago
11

2 5/8 - 6 1/4 + 8 1/2

Mathematics
1 answer:
jek_recluse [69]3 years ago
3 0

Answer:

2 5/8 - 6 1/4 + 8 1/2\

= 227 /8

Step-by-step explanation:

hope this helps :)

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Which expression represent the approximate length of BC HEEEEELLLLPPP!!!
Bumek [7]

Answer:

A

Step-by-step explanation:

Using the Sine rule in Δ ABC

\frac{sinA}{BC} = \frac{sin38}{AB} , substitute values

\frac{sin66}{BC} = \frac{sin38}{3} ( cross- multiply )

BC sin38° = 3 sin66° ( divide both sides by sin38° )

BC = \frac{3sin66}{sin38} → A

5 0
3 years ago
A random sample of 30 households was selected as part of a study on electricity usage, and the number of kilowatt-hours (kWh) wa
grin007 [14]

Answer:

The 98% confidence interval for the mean usage in the March quarter of 2006, in kWh, was (333.87, 416.13).

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 30 - 1 = 29

98% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 29 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.98}{2} = 0.99. So we have T = 2.462

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 2.462\frac{91.5}{\sqrt{30}} = 41.13

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 375 - 41.13 = 333.87 kWh

The upper end of the interval is the sample mean added to M. So it is 375 + 41.13 = 416.13 kWh

The 98% confidence interval for the mean usage in the March quarter of 2006, in kWh, was (333.87, 416.13).

6 0
3 years ago
10x +2y = 64
Musya8 [376]

Answer:

<h2>x = 4, y = 12 → (4, 12)</h2>

Step-by-step explanation:

\left\{\begin{array}{ccc}10x+2y=64&\text{multiply both sides by 2}\\3x-4y=-36\end{array}\right\\\underline{+\left\{\begin{array}{ccc}20x+4y=128\\3x-4y=-36\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad23x=92\qquad\text{divide both sides by 2}\\.\qquad x=4\\\\\text{put the value of x to the first equation:}\\\\10(4)+2y=64\\40+2y=64\qquad\text{subtract 40 from both sides}\\2y=24\qquad\text{divide both sides by 2}\\y=12

5 0
4 years ago
Find the volume of the solid under the surface z = 5x + 9y 2 and above the region bounded by x = y 2 and x = y 3.
Yanka [14]
Call the region in the x-y plane, bounded by x=y^2 and x=y^3, \mathcal D. Then the volume under the given surface is

\displaystyle\iint_{\mathcal D}(5x+9y)\,\mathrm dA=\int_{y=0}^{y=1}\int_{x=y^3}^{x=y^2}(5x+9y)\,\mathrm dx\,\mathrm dy=\frac{83}{140}
8 0
3 years ago
Someone who me with this entire page please
olganol [36]
Hopefully this helps!!!

8 0
3 years ago
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