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3 years ago

Help me!! please. find the number of each numbered angle

Mathematics

1 answer:

Whitepunk [10]3 years ago

5 0

Answer:

Angle(1) = 60°

Angle(2) = 50°

Angle(3) = 70°

Explanation:

I'm much too tired to explain it rn but still if you really need the explanation, I'll give it to you in the comments section :)

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If 9 = 5 + 4 and 5 + 4 = 3 + 6, then 9 = 3 + 6. This is an example of which property?

Anna007 [38]

B. Transitive Property

8 0

3 years ago

Read 2 more answers

I need help quick! <br> Q: The measure of

Citrus2011 [14]

Answer:

x=34

Step-by-step explanation:

132-64=

68/2=

PLEASE MAKE ME BRAINLIEST

7 0

3 years ago

A state legislator wants to determine whether his voters' performance rating (0 - 100) has changed from last year to this year.

ra1l [238]

Answer:

Step 1 of 4

Point estimate for the population mean of the paired differences = -8.2

Step 2 of 4

Sample standard deviation of the paired differences = 16.116244

Step 3 of 4

Margin of Error = ±9.326419

Step 4 of 4

90% Confidence interval = (-17.5, 1.1)

Step-by-step explanation:

The ratings from last year and this year are given in table as

Rating (last year) | x1 | 87 67 68 75 59 60 50 41 75 72

Rating (this year) | x2| 85 52 51 53 50 52 80 44 48 57

Difference | x2 - x1 | -2 -15 -17 -22 -9 -8 30 3 -27 -15

Step 1 of 4

Mean = (Σx)/N = (-82/10) = -8.2 to 1 d.p.

Step 2 of 4

Standard deviation for the sample

= √{[Σ(x - xbar)²]/(N-1)} = 16.116244392951 = 16.116244 to 6 d.p.

Step 3 of 4

Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Sample Mean = -8.2

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Critical value will be obtained using the t-distribution. This is because there is no information provided for the population standard deviation.

To find the critical value from the t-tables, we first find the degree of freedom and the significance level.

Degree of freedom = df = n - 1 = 10 - 1 = 9.

Significance level for 90% confidence interval

= (100% - 90%)/2 = 5% = 0.05

t (0.05, 9) = 1.83 (from the t-tables)

Standard error of the mean = σₓ = (σ/√n)

σ = standard deviation of the sample = 16.116244

n = sample size = 10

σₓ = (16.116244/√10) = 5.0964038367

Margin of Error = (Critical value) × (standard Error of the mean) = 1.83 × 5.0964038367 = 9.3264190212 = 9.326419 to 6 d.p.

Step 4 of 4

90% Confidence Interval = (Sample mean) ± (Margin of Error)

CI = -8.2 ± (9.326419)

90% CI = (-17.5264190212, 1.1264190212)

90% Confidence interval = (-17.5, 1.1)

Hope this Helps!!!

4 0

3 years ago

A simulation was conducted using 10 fair six-sided dice, where the faces were numbered 1 through 6. respectively. All 10 dice we

kompoz [17]

Answer:

C) a sample distribution of a sample mean with n = 10

$\mu_{{\overline}{X}} = 3.5$

and $\sigma_{{\overline}{Y}} = 0.38$

Step-by-step explanation:

Here, the random experiment is rolling 10, 6 faced (with faces numbered from 1 to 6) fair dice and recording the average of the numbers which comes up and the experiment is repeated 20 times.So, here sample size, n = 20 .

Let,

$X_{ij}$ = The number which comes up on the ith die on the jth trial.

∀ i = 1(1)10 and j = 1(1)20

Then,

$E(X_{ij})$ = $\frac {1 + 2 + 3 + 4 + 5 + 6}{6}$

= 3.5 ∀ i = 1(1)10 and j = 1(1)20

and,

$E(X^{2}_{ij}$ = $\frac {1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2}}{6}$

= $\frac {1 + 4 + 9 + 16 + 25 + 36}{6}$

= $\frac {91}{6}$

$\simeq$ 15.166667

so, $Var(X_{ij}$ = $(E(X^{2}_{ij} - {(E(X_{ij})}^{2})$

$\simeq 15.166667 - 3.5^{2}$

= 2.91667

and $\sigma_{X_{ij}}$ = $\sqrt {2.91667}[/tex [tex]\simeq 1.7078261036$

Now we get that,

$Y_{j} = \frac {\sum_{j = 1}^{20}X_{ij}}{20}$

We get that $Y_{j}'s$ are iid RV's ∀ j = 1(1)20

Let, ${\overline}{Y} = \frac {\sum_{j = 1}^{20}Y_{j}}{20}$

So, we get that $E({\overline}{Y}) = E(Y_{j})$

= $E(X_{ij}$ for any i = 1(1)10

= 3.5

and,

$\sigma_{({\overline}{Y})} = \frac {\sigma_{Y_{j}}}{\sqrt {20}} = \frac {\sigma_{X_{ij}}}{\sqrt {20}} = \frac {1.7078261036}{\sqrt {20}} [tex]\simeq 0.38$

Hence, the option which best describes the distribution being simulated is given by,

C) a sample distribution of a sample mean with n = 10

$\mu_{{\overline}{X}} = 3.5$

and $\sigma_{{\overline}{Y}} = 0.38$

6 0

3 years ago

Which of the following proportionally statements is correct?

yulyashka [42]

Answer:

$\frac{AC}{BC}=\frac{DF}{EF}$

Step-by-step explanation:

The figure given has two similar triangles: ΔABC and ΔDEF. Although the triangles are similar, their orientation is different and ΔDEF is flipped. Since the triangles are similar, their side lengths are proportional to each other. Given the orientation of the triangles, we can still see that the diagonal (hypotenuse) for the larger triangle is AC and the smaller is DF. The only answer that matches these up proportionally is the third one. Looking at the second side, BC, we can see that this matches up to the longer leg of EF on the smaller triangle. Final answer being:

$\frac{AC}{BC}=\frac{DF}{EF}$

6 0

4 years ago