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Dimas [21]
3 years ago
13

Are these triangles similar Please hurry!

Mathematics
2 answers:
mezya [45]3 years ago
4 0

Answer:

no they are not

Step-by-step explanation:

mariarad [96]3 years ago
4 0

<em>Answer:</em>

<em>ΔQRS ≠ ΔTVW</em>

<em>Step-by-step explanation:</em>

<em>∡Q ≠ ∡T</em>

<em>∡R ≠ ∡V => ΔQRS ≠ ΔTVW</em>

<em>∡S ≠ ∡W</em>

<em />

You might be interested in
The Table below gives you The population of Texas since 1970. Find the average rate if change from 1970 to 1090
ruslelena [56]

Answer:

Average rate of change = 0.29 million per year

Step-by-step explanation:

Average rate of change in the population from year 1970 to year 1900 is given by the formula,

Average rate of change = \frac{\text{Population in year 1990-Population in year 1970}}{1990-1970}

                                        = \frac{17-11.2}{1990-1970}

                                        = \frac{5.8}{20}

                                        = 0.29 million per year

Therefore, average rate of change in the population from 1970 to 1990 will be 0.29 million per year.                        

8 0
3 years ago
2. The average score of all golfers for a particular course has a mean of 75 and a standard deviation of 4.5. Suppose we going t
stiv31 [10]

Answer:

P(\bar X>76)

And we can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And if we find the z score for 76 we got:

z =\frac{76-75}{\frac{4.5}{\sqrt{81}}}= 2

And if we use the complement rule and the normal standard distribution or excel we got:

P(Z>2) = 1-P(Z

And we can find this using the ti 84 with the following code:

1-normalcdf(-1000, 76, 75, 0.5 )

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Let X the random variable who represent the score of golfers. We select a sampel size of 81.

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And we want to find this probability:

P(\bar X>76)

And we can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And if we find the z score for 76 we got:

z =\frac{76-75}{\frac{4.5}{\sqrt{81}}}= 2

And if we use the complement rule and the normal standard distribution or excel we got:

P(Z>2) = 1-P(Z

And we can find this using the ti 84 with the following code:

1-normalcdf(-1000, 76, 75, 0.5 )

6 0
3 years ago
Graph the equation of the line y=3x+4 and use it for your reference to find: Quadrants in which this graph is in. (Use the numbe
Dahasolnce [82]

Answer:

Quadrants: 1, 3, 4. (See explanation for further details)

Step-by-step explanation:

The graphic of the linear function is included below as attachment. The line covers the following quadrants: 1, 3, 4

3 0
3 years ago
4.65 rounded to the nearest whole
DIA [1.3K]

Answer:

5

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Would appreciate the help
sergejj [24]

Answer:

x = 13

Step-by-step explanation:

Hope it helps you!........

8 0
3 years ago
Read 2 more answers
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