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2 years ago

1-07 What is the length of the marked portion of each line segment? Copy the

Mathematics

1 answer:

VikaD [51]2 years ago

5 0

Answer:

A. 25 B. 45 C. 30

Step-by-step explanation:

75 ÷ 3 = 25

25 × 1 = 25

75 ÷ 5 = 15

15 × 3 = 45

50 ÷ 5 = 10

10 × 3 = 30

You might be interested in

Find the volume of the figure below. 5 cm 11 cm 66 cm 82.5 cm 132 cm 165 cm

stealth61 [152]

Answer:

A) 66cm³

Step-by-step explanation:

Using the Pythagorean Theorem, the base of the triangle is 4. The area of a triangle is 1/2 base times height. Then you multiply times 11 to get the volume.

1/2(3)(4)(11)=66

5 0

3 years ago

Graph y=-6(3)^x please do the table and the graph 20 POINTS!

Margarita [4]

Answer:

Explanation to the answer below... (Answers in the Explanation)

Step-by-step explanation:

We Know that our rule is y = $y = -6(3)^{x}$ , which means we can use this rule to find y from x. Since we don't know what X, is, we have to assume numbers like 1, to 7. And there is no space to write on the graph, I'll put a virtual copy of a graph.

1st Point,

$y = -6(3)^{x}\\y = -6(3)^{1}\\y = -6(3)\\y = -18\\x = 1, y = -18$

2nd, point

$y = -6(3)^x\\ y = -6(3)^2\\y = -6(9)\\y = -54\\x = 2, y = -54$

3rd Point

$y = -6(3)^x\\y = -6(3)^3\\y = -6(27)\\y = -162\\x = 3, y = -162$

4th Point

$y = -6(3)^4\\y = -6(81)\\y = -486\\x = 4, y = -486\\$

5th Point

$y = -6(3)^5\\y = -6(243)\\y = -1458\\\x = 5, y = -1458$

6th Point

$y = -6(3)^x\\y = -6(3)^6\\y = -6(729)\\y = -4,374\\x = 6, y = -4,374\\$

7th Point (Last Point)

$y = -6(3)^x\\y = -6(3)^7\\y = -6(2,187)\\y. = -13,122\\x = 7, y = -13,122$

Let's now plot them on the Graph, the values are so big, the graph might look a little odd.

Always remember that we can continue the points infinitely, and there is no stop, The Graph is attached below...

4 0

2 years ago

3 Which expression is closest in value to 5.66?

Free_Kalibri [48]

Let's consider the standard and usually used value of irrational numbers given in the question.

It is asked to see which is closed to 5.66

So, let's see the options:

A) GiveN:

$\dfrac{\pi}{2} + \pi$

Let consider π = 3.14

$\dfrac{3.14}{2} + 3.14$

$1.57 + 3.14$

$4.71 \: (approx.)$

B) GiveN:

$2\pi - 1$

Here also, let's consider the value be 3.14

$2 \times 3.14 - 1$

$6.28 - 1$

$5.28 \: (approx.)$

C) GiveN:

$4 \sqrt{2}$

Let's consider the value of √2 be 1.414

$4 \times 1.414$

$5.6516 \: (approx.)$

D) GiveN:

$8 - \sqrt{3}$

Let's consider the value of √3 be 1.732

$8 - 1.732$

$6.268 \: (approx.)$

So, we can see that we have got the approximate value of all the numericals, and the closest to 5.66 is 5.6516 which is the answer of Option C.

So, Correct answer is C

#CarryOnLearning

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6 0

3 years ago

A researcher used a sample of n = 60 individuals to determine whether there are any preferences among six brands of pizza. Each

Blizzard [7]

Answer:

1) χ² ≥ 11.07

2) Goodness of fit test, df: χ² $_{3}$

Independence test, df: χ² $_{1}$

The goodness of fit test has more degrees of freedom than the independence test.

3) $e_{females.} = 80$

4) H₀: $P_{ij}= P_{i.} * P_{.j}$ ∀ i= 1, 2, ..., r and j= 1, 2, ..., c

5) χ² $_{6}$

Step-by-step explanation:

Hello!

The researcher took a sample of n=60 people and made them taste proof samples of six different brands of pizza and choose their favorite brand, their choose was recorded. So the study variable is the following:

X: favorite pizza brand, categorized in brand 1, brand 2, brand 3, brand 4, brand 5 and brand 6.

The Chi-square goodness of fit test is done with the following statistic:

χ²= ∑ $\frac{(O_i-E_i)^2}{E_i}$ ≈χ² $_{k-1}$

Where k represents the number of categories of the study variable. In this example k= 6.

Remember, the rejection region for the Chi-square tests of "goodnedd of fit", "independence", and "homogeneity" is allways one-tailed to the right. So you will only have one critical value.

χ² $_{k-1; 1 - \alpha }$

χ² $_{6-1; 1 - 0.05 }$

χ² $_{5; 0.95 }$ = 11.070

This means thar the rejection region is χ² ≥ 11.07

If the Chi-Square statistic is equal or greather than 11.07, then you reject the null hypothesis.

The statistic for the goodness of fit is:

χ²= ∑ $\frac{(O_i-E_i)^2}{E_i}$ ≈χ² $_{k-1}$

Degrees of freedom: χ² $_{k-1}$

In the example: k= 4 (the variable has 4 categories)

χ² $_{4-1}$ = χ² $_{3}$

The statistic for the independence test is:

χ²= ∑∑ $\frac{(O_ij-E_ij)^2}{E_ij}$ ≈χ² $_{(r-1)(c-1)}$ ∀ i= 1, 2, ..., r & j= 1, 2, ..., c

If the information is in a contingency table

r= represents the total of rows

c= represents the total of columns

In the example: c= 2 and r= 2

Degrees of freedom: χ² $_{(r-1)(c-1)}$

χ² $_{(2-1)(2-1)}$ = χ² $_{1}$

The goodness of fit test has more degrees of freedom than the independence test.

To calculate the expected frecuencies for the independence test you have to use the following formula.

$e_{ij} = n * P_i. * P_.j = n * \frac{o_i.}{n} * \frac{o_.j}{n}$

Where o_i. represents the total observations of the i-row, o_.j represents the total of observations of the j-column and n is the sample size.

Now, this is for the expected frequencies in the body of the contingency table, this means the observed and expected frequencies for each crossing of categories is not the same.

On the other hand, you would have the totals of each category and population in the margins of the table (subtotals), this is the same when looking at the observed frequencies and the expected frequencies. Wich means that the expected frequency for the total of a population is the same as the observed frequency of said population. A quick method to check if your calculations of the expected frequencies for one category/population are correct is to add them, if the sum results in the subtotal of that category/population, it means that you have calculated the expected frequencies correctly.

The expected frequency for the total of females is 80

Using the formula:

(If the females are in a row) $e_{females.} = 100 * \frac{80}{100} * \frac{0}{100}$

$e_{females.} = 80$

There are two ways of writing down a null hypothesis for the independence test:

Way 1: using colloquial language

H₀: The variables X and Y are independent

Way 2: Symbolically

H₀: $P_{ij}= P_{i.} * P_{.j}$ ∀ i= 1, 2, ..., r and j= 1, 2, ..., c

This type of hypothesis follows from the definition of independent events, where if we have events A and B independent of each other, the probability of A and B is equal to the product of the probability of A and the probability of B, symbolically: P(A∩B) = P(A) * P(B)

In this example, you have an independence test for two variables.

Variable 1, has 3 categories

Variable 2, has 4 categories

To follow the notation, let's say that variable 1 is in the rows and variable 2 is in the columns of the contingency table.

The statistic for this test is:

χ²= ∑∑ $\frac{(O_ij-E_ij)^2}{E_ij}$ ≈χ² $_{(r-1)(c-1)}$ ∀ i= 1, 2, ..., r & j= 1, 2, ..., c

In the example: c= 3 and r= 4

Degrees of freedom: χ² $_{(r-1)(c-1)}$

χ² $_{(3-1)(4-1)}$ = χ² $_{6}$

I hope you have a SUPER day!

4 0

3 years ago

Write the equation of a line that is perpendicular to y=-2x/7+9 and that passes through the point (4,−6)

Shtirlitz [24]

Answer:

y = (7/2)x -20

Step-by-step explanation:

The given line is in slope-intercept form, so you can read its slope from the equation.

y = mx + b . . . . . m is the slope; b is the y-intercept

y = -(2/7)x + 9 . . . . . . has slope -2/7

The perpendicular line will have a slope that is the negative reciprocal of this, so will be ...

m = -1/(-2/7) = 7/2

We can use this and the given point to write the equation in point-slope form.

y = m(x -h) +k . . . . . . line with slope m through point (h, k)

We have m = 7/2, (h, k) = (4, -6) so the equation is ...

y = (7/2)(x -4) -6

y = (7/2)x -20

4 0

3 years ago