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3 years ago

PLS HELP MEEEI NEED ANSWERS ASAP ASAP

Mathematics

1 answer:

Brums [2.3K]3 years ago

6 0

Answer:

Associative.

Step-by-step explanation:

In addition and multiplication, it does not matter which two numbers you put together first. On the left aide 6 is associated with 3. On the right side 3 is grouped with 7. It doesn't matter where 3 goes. You still get the same answer.

(6 + 3) + 7 = 6 + (3 + 7)

9 + 7 = 6 + 10

16 = 16

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Sunspots have been observed for many centuries. Records of sunspots from ancient Persian and Chinese astronomers go back thousan

Lunna [17]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ $\leq$ 41

Alternate Hypothesis, $H_A$ : $\mu$ > 41

(b) The value of z test statistics is 1.08.

(c) We conclude that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41.

Step-by-step explanation:

We are given that in a random sample of 40 such periods from Spanish colonial times, the sample mean is x¯ = 47.0. Previous studies of sunspot activity during this period indicate that σ = 35.

It is thought that for thousands of years, the mean number of sunspots per 4-week period was about µ = 41.

Let $\mu$ = <u><em>mean sunspot activity during the Spanish colonial period.</em></u>

(a) Null Hypothesis, $H_0$ : $\mu$ $\leq$ 41 {means that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41}

Alternate Hypothesis, $H_A$ : $\mu$ > 41 {means that the mean sunspot activity during the Spanish colonial period was higher than 41}

The test statistics that would be used here <u>One-sample z test statistics</u> as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean = 47

σ = population standard deviation = 35

n = sample of periods from Spanish colonial times = 40

So, <em><u>the test statistics</u></em> = $\frac{47-41}{\frac{35}{\sqrt{40} } }$

= 1.08

(b) The value of z test statistics is 1.08.

(c) <u>Now, the P-value of the test statistics is given by;</u>

P-value = P(Z > 1.08) = 1 - P(Z < 1.08)

= 1 - 0.8599 = <u>0.1401</u>

Since, the P-value of the test statistics is higher than the level of significance as 0.1401 > 0.05, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41.

8 0

4 years ago

The ratio of boys to girls in Ella’s grade is 3 to 4. The ratio of boys to girls in Minh’s grade is 4 to 5. Each grade has a tot

Alex_Xolod [135]

Answer:

12 boys

Step-by-step explanation:

I believe the answer is 12 boys. I hope this helps!

8 0

3 years ago

Pls Help This is Due Soon (4/27): The following ordered pairs from a linear function: (-5,-7), (-2, 2), (3, 17), (7, 29), аnd (1

lesantik [10]

-7 ur welcome

can you please mark me as brainlist

4 0

3 years ago

Read 2 more answers

Keith drove 436 miles using 16 gallons of gas at this rate how many miles would he drive using 14 gallons of gas

KengaRu [80]

This Question is related to Direct Proportionality

Here : The Number of Miles Driven is Directly Proportional to Number of Gallons of Gas

Number of Miles ∝ Gallons of Gas

Given Keith with 16 Gallons of Gas Drove 436 Miles

⇒ Keith with 1 Gallon of Gas would Drive - $\frac{436}{16}$ Miles

⇒ Keith with 1 Gallon of Gas would Drive : 27.25 Miles

⇒ Keith with 14 Gallons of Gas would Drive : 14 × 27.25 Miles

⇒ Keith with 14 Gallons of Gas would Drive : 381.5 Miles

8 0

3 years ago

21. Who is closer to Cameron? Explain.

pickupchik [31]

Problem 21

<h3>Answer: Jamie is closer</h3>

-----------------------

Explanation:

A = Arthur's location = (20,35)
J = Jamie's location = (45,20)
C = Cameron's location = (65,40)

To find out who's closer to Cameron, we need to compute the segment lengths AC and JC. Then we pick the smaller of the two lengths.

We use the distance formula to find each length

Let's find the length of AC.

$A = (x_1,y_1) = (20,35)\\\\C = (x_2,y_2) = (65,40)\\\\d = \text{Distance from A to C} = \text{length of segment AC}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(20-65)^2 + (35-40)^2}\\\\d = \sqrt{(-45)^2 + (-5)^2}\\\\d = \sqrt{2025 + 25}\\\\d = \sqrt{2050}\\\\d \approx 45.2769257\\\\$

The distance from Arthur to Cameron is roughly 45.2769257 units.

Let's repeat this process to find the length of segment JC

$J = (x_1,y_1) = (45,20)\\\\C = (x_2,y_2) = (65,40)\\\\d = \text{Distance from J to C} = \text{length of segment JC}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(45-65)^2 + (20-40)^2}\\\\d = \sqrt{(-20)^2 + (-20)^2}\\\\d = \sqrt{400 + 400}\\\\d = \sqrt{800}\\\\d \approx 28.2842712\\\\$

Going from Jamie to Cameron is roughly 28.2842712 units

We see that segment JC is shorter than AC. Therefore, Jamie is closer to Cameron.

=================================================

Problem 22

<h3>Answer: Arthur is closest to the ball</h3>

-----------------------

Explanation:

We have these key locations:

A = Arthur's location = (20,35)
J = Jamie's location = (45,20)
C = Cameron's location = (65,40)
B = location of the ball = (35,60)

We'll do the same thing as we did in the previous problem. This time we need to compute the following lengths:

These segments represent the distances from a given player to the ball. Like before, the goal is to pick the smallest of these segments to find out who is the closest to the ball.

The steps are lengthy and more or less the same compared to the previous problem (just with different numbers of course). I'll show the steps on how to get the length of segment AB. I'll skip the other set of steps because there's only so much room allowed.

$A = (x_1,y_1) = (20,35)\\\\B = (x_2,y_2) = (35,60)\\\\d = \text{Distance from A to B} = \text{length of segment AB}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(20-35)^2 + (35-60)^2}\\\\d = \sqrt{(-15)^2 + (-25)^2}\\\\d = \sqrt{225 + 625}\\\\d = \sqrt{850}\\\\d \approx 29.1547595\\\\$

Segment AB is roughly 29.1547595 units.

If you repeated these steps, then you should get these other two approximate segment lengths:

JB = 41.2310563

CB = 36.0555128

-------------

So in summary, we have these approximate segment lengths

AB = 29.1547595
JB = 41.2310563
CB = 36.0555128

Segment AB is the smallest of the trio, which therefore means Arthur is closest to the ball.

3 0

3 years ago