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2 years ago

I need help with this question

1 answer:

4 0

We use a chi-square test of homogeneity, as we have a single categorical variable measured within two different populations.

The requirements for this test are that we have two simple random samples (this is clearly satisfied, given that we are sampling all students who take the class that year) and that the expected value for each cell, defined as the cell's row sum multiplied by the column sum divided by the sum of all values in the dataset, is greater than 5. This is satisfied simply by testing for each value in the table.

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Which inequality does the number line represent? A) x + 3 > −7 B) x − 3 < −4 C) x − 3 ≤ −7 D) x − 3 ≥ −7

Vesnalui [34]

Hey love where is the number line to answer this question?

4 0

3 years ago

Find area of the shaded region.

Monica [59]

Answer:

The area of the shaded region is about 38.1 square centimeters.

Step-by-step explanation:

We want to find the area of the shaded region.

To do so, we can first find the area of the sector and then subtract the area of the triangle from the sector.

The given circle has a radius of 6 cm.

And the given sector has a central angle of 150°.

The area for a sector is given by the formula:

$\displaystyle A=\pi r^2\cdot \frac{\theta}{360^\circ}$

In this case, r = 6 and θ = 150°. Hence, the area of the sector is:

$\displaystyle \begin{aligned}A&=\pi(6)^2\cdot \frac{150}{360}\\ &=36\pi\cdot \frac{5}{12}\\&=3\pi \cdot 5\\&=15\pi \text{ cm}^2\end{aligned}$

Now, we can find the area of the triangle. We can use an alternative formula:

$\displaystyle A=\frac{1}{2}ab\sin(C)$

Where a and b are the side lengths, and C is the angle between them.

Both side lengths of the triangle are the radii of the circle. So, both side lengths are 6.

And the angle C is 150°. Hence, the area of the triangle is:

$\displaystyle A=\frac{1}{2}(6)(6)\sin(150)=18\sin(150)$

The area of the shaded region is equivalent to the sector minus the triangle:

$A_{\text{shaded}}=A_{\text{sector}}-A_{\text{triangle}}$

Therefore:

$A_{\text{shaded}}=15\pi -18\sin(150)$

Use a calculator:

$A_{\text{shaded}}=38.1238...\approx 38.1\text{ cm}^2$

The area of the shaded region is about 38.1 square centimeters.

6 0

3 years ago

Learning Task 3

ASHA 777 [7]

Answer:

1. The sides are: 38.5, 39.5, 40.5 and 41.5

2. Presently, Inzo is 15 years old and Miz is 10 years old

3. $Distance = 3682km$

4. Possible pairs are: (12, 14), (16, 18) and (20, 24)

5. $L < 38m$

Step-by-step explanation:

The question has lots of missing details. See the comment section for complete question

Solving (a):

Given

Shape: Quadrilateral

Sides: Consecutive

$Perimeter = 160$

Let one of the sides be x. The other sides will be x + 1, x + 2 and x + 3.

So, the perimeter is:

$Perimeter = x + x + 1 + x + 2 + x + 3$

This gives:

$160 = x + x + 1 + x + 2 + x + 3$

Collect Like Terms

$160 -1-2-3= x+x+x+x$

$154= 4x$

Divide both sides by 4

$38.5 = x$

$x =38.5$

So, the sides are: 38.5, 39.5, 40.5 and 41.5

Solving (b):

Given

Presently: $Inzo = 10 + Miz$

Five years ago: $Miz - 5= \frac{1}{3} * (Inzo-5)$ --- (Missing from the question)

Required: Determine their present ages

Substitute $10 + Miz$ for Inzo in the second equation

$Miz - 5 = \frac{1}{3} * (10 + Miz - 5)$

$Miz - 5 = \frac{1}{3} * (Miz + 5)$

Multiply both sides by 3

$3Miz - 15 = Miz + 5$

Collect Like Terms

$3Miz -Miz = 15 + 5$

$2Miz= 20$

Divide both sides by 2

$Miz = 10$

Substitute 10 for Miz in $Inzo = 10 + Miz$

$Inzo = 10 + 5$

$Inzo = 15$

So presently, Inzo is 15 years old and Miz is 10 years old

Solving (3):

Given

Express Train:

$Speed = 100kph$

$Time = t$

Local Train:

$Speed = 45kph$

$Time = t + 45$

Required: Determine the distance between A and B

Distance is calculated as:

$Distance = Speed * Time$

For the express train:

$Distance = 100 * t$

For the local train

$Distance = 45 * (45 + t)$

Distance between both stations are equal; So,

$100 * t = 45 * (45 + t)$

$100 t = 2025 + 45t$

Collect Like Terms

$100 t -45t= 2025$

$55t= 2025$

Divide by 55

$t = 36.82$

Substitute 36.82 for t in $Distance = 100 * t$

$Distance = 100 * 36.82$

$Distance = 3682km$

Solving (4):

Let the even numbers be: n and n + 2.

Sum greater than 24: $n + n + 2 > 24$

Sum less than 60: $n + n + 2 < 60$

So, we have:

$n + n + 2 > 24$

$2n + 2 > 24$

Collect Like Terms

$2n > 24-2$

$2n > 22$

$n > 11$

$n + n + 2 < 60$

$2n + 2< 60$

Collect Like Terms

$2n < 60-2$

$2n < 58$

Divide by 2

$n$

So, we have:

$n > 11$ and $n < 29$

So, possible pairs are: (12, 14), (16, 18) and (20, 24)

Solving (5):

Given

$Perimeter < 152m$

Required: Find the possible lengths

Let the length be L.

Perimeter is calculated as:

$Perimeter =4 * L$

$Perimeter =4 L$

So, we have:

$4L < 152m$

Divide through by 4

$L < 38m$

Hence, length is less than 38m (e.g. 37m)

5 0

2 years ago

They’re 80 skittles in a bag 20% of them are read how many skittles a red

Nutka1998 [239]

Answer: 16

Step-by-step explanation: 80x20=1600

Put the decimal place where it belongs

80% of 20=16

Hope this helps!

7 0

3 years ago

What is the nearest whole number rounded to 67400

viktelen [127]

67401 or 67400 or 67399

8 0

3 years ago