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Vaselesa [24]
3 years ago
5

HELPPPPPPPPPPPPPPPPPP

Mathematics
1 answer:
igor_vitrenko [27]3 years ago
6 0
It’s 2 units I think .
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X+4=566666<br> uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
melisa1 [442]

Answer:

x=566662 or x=52

Step-by-step explanation:

Im going to answer this multiple times, in case the extra 6s are a typo.

<em>Answer 1:</em>

x+4=566666

  -4   -4

x=566662

<em>Answer 2:</em>

x+4=56

x=52

5 0
4 years ago
Please help the picture is
asambeis [7]

The walls are vertical so the angle opposite X would be a right angle which is 90 degrees.

X would be 180 - 90 - 35 = 55 degrees

X and 2y - 5 are complementary angles which add together to equal 90:

2y - 5 + 55 = 90

Simplify:

2y +50 = 90

Subtract 50 from both sides:

2y = 40

Divide both sides by 2:

y = 20

7 0
3 years ago
Read 2 more answers
The temperature of a certain solution is estimated by taking a large number of independent measurements and averaging them. The
Mademuasel [1]

Answer:

(a) The 95% confidence interval for the temperature is (36.80°C, 37.20°C).

(b) The confidence level is 68%.

(c) The necessary assumption is that the population is normally distributed.

(d) The 95% confidence interval for the temperature if 10 measurements were made is (36.93°C, 37.07°C).

Step-by-step explanation:

Let <em>X</em> = temperature of a certain solution.

The estimated mean temperature is, \bar x=37^{o}C.

The estimated standard deviation is, s=0.1^{o}C.

(a)

The general form of a (1 - <em>α</em>)% confidence interval is:

CI=SS\pm CV\times SD

Here,

SS = sample statistic

CV = critical value

SD = standard deviation

It is provided that a large number of independent measurements are taken to estimate the mean and standard deviation.

Since the sample size is large use a <em>z</em>-confidence interval.

The critical value of <em>z</em> for 95% confidence interval is:

z_{0.025}=1.96

Compute the confidence interval as follows:

CI=SS\pm CV\times SD\\=37\pm 1.96\times 0.1\\=37\pm0.196\\=(36.804, 37.196)\\\approx (36.80^{o}C, 37.20^{o}C)

Thus, the 95% confidence interval for the temperature is (36.80°C, 37.20°C).

(b)

The confidence interval is, 37 ± 0.1°C.

Comparing the confidence interval with the general form:

37\pm 0.1=SS\pm CV\times SD

The critical value is,

CV = 1

Compute the value of P (-1 < Z < 1) as follows:

P(-1

The percentage of <em>z</em>-distribution between -1 and 1 is, 68%.

Thus, the confidence level is 68%.

(c)

The confidence interval for population mean can be constructed using either the <em>z</em>-interval or <em>t</em>-interval.

If the sample selected is small and the standard deviation is estimated from the sample, then a <em>t</em>-interval will be used to construct the confidence interval.

But this will be possible only if we assume that the population from which the sample is selected is Normally distributed.

Thus, the necessary assumption is that the population is normally distributed.

(d)

For <em>n</em> = 10 compute a 95% confidence interval for the temperature as follows:

The (1 - <em>α</em>)% <em>t</em>-confidence interval is:

CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}

The critical value of <em>t</em> is:

t_{\alpha/2, (n-1)}=t_{0.025, 9}=2.262

*Use a <em>t</em>-table for the critical value.

The 95% confidence interval is:

CI=37\pm 2.262\times \frac{0.1}{\sqrt{10}}\\=37\pm 0.072\\=(36.928, 37.072)\\\approx (36.93^{o}C, 37.07^{o}C)

Thus, the 95% confidence interval for the temperature if 10 measurements were made is (36.93°C, 37.07°C).

3 0
3 years ago
An oil company fills of a tank in hour. At this rate, which expression can be used to determine how long it will take for the ta
monitta
Is there anymore to the question? I need some more information to be able to help
6 0
3 years ago
Read 2 more answers
8th grade math<br>m&lt;T=<br>m&lt;F=<br><br>pls explain I have no clue what I'm doing​
pishuonlain [190]

180

-90

--------

90

-----

2

=

45

both are 45 degrees.

3 0
3 years ago
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