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Sedaia [141]
3 years ago
15

How do you do questions 3 and 4 and explain if you can.

Mathematics
1 answer:
ioda3 years ago
6 0
Well for question 3 you could use algebra. if there are 4568 peaches and mangoes together and 1654 more peaches than mangoes. mangoes qill be our x and peaches will be x + 1654 because there are 1654 more peaches than mangoes. so since peaches + mangoes = 4568, x + 1654(peaches) + x(mangoes) = 4568, now we have to do the algebra. x + 1654 + x = 4568 , first collect like terms, x + x = 2x. 2x +1654 = 4568 now we subtract 1654 - 1654 and 4568 - 1654. 2x = 2914. now we do 2 divided by 2 and 2914 divided by 2. our answer is x = 1457, that's how many mangoes there are, now we do 1457 + 1654 to find out how many peaches there are, there are 3111 peaches. now let's check to see if we're right, 1457 + 3129 = 4568
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I will give brainliest to people who answer ALL 5 questions.
amid [387]
Here ya go. Hope this helps. Have a great day

8 0
3 years ago
In order to ensure efficient usage of a server, it is necessary to estimate the mean number
juin [17]

Answer:

a. [36.19;39.21]

b. Reject the null hypothesis. The population mean of users that are connected at the same time is greater than 35.

Step-by-step explanation:

Hello!

Your study variable is,

X: "number of users of one server at a time"

The objective is to estimate the mean, for this, a sample of n=100 times was taken and the standard deviation S= 9.2 and the sample mean is X[bar]= 37.7 were calculated.

You need to study the population mean, for this you need your variable to have at least normal distribution. Since you don't have information about its distribution, but the sample is big enough (n≥30) you can apply the Central Limit Theorem and approximate the distribution of the sample mean X[bar] to normal:

X[bar]≈N(μ;σ²/n)

a. With this approximation, you can construct the 90% Confidence Interval using the approximate Z

[X[bar] ± Z_{1-\alpha /2} * S/√n]

Z_{1-\alpha /2} = Z_{0.95} = 1.64

[37.7± 1.64* 9.2/√100]

[36.19;39.21]

b. You need to test if the population mean is greater than 35 with a level of significance of 1%.

The hypothesis is:

H₀: μ ≤ 35

H₁: μ > 35

α: 0.01

This is a one-tailed test so you have only one critical level (right tail):

Z_{1\alpha } = Z_{0.99} = 2.33

This means that if the value of the calculated statistic is equal or greater than 2.33 you will reject the null Hypothesis.

If the value is less than 2.33 you will support the null hypothesis.

The statistic is:

Z=<u> X[bar] - μ </u>= <u> 37.7 - 35 </u> = 2.93

       S/√n           9.2/10

The value 2.93 > 2.33, so you reject the null hypothesis. This means that the population mean of users that are connected at the same time is greater than 35.

<u><em>Note: </em></u><em>To make the decision using the interval calculated on a), the hypothesis should have been two-tailed and the confidence and significance levels complementary.</em>

I hope it helps!

7 0
3 years ago
A rectangle has a length of (x+10) centimeters and a width of x centimeters. If the perimeter is 32 centimeters, what is the len
I am Lyosha [343]

Given:

Length of rectangle = (x+10) cm

Width of the rectangle = x cm

Perimeter = 32 cm

To find:

The length of the rectangle.

Solution:

We know that,

Perimeter=2(l+w)

Where, l is length and w is width.

Substituting the values, we get

32=2((x+10)+x)

32=2(2x+10)

32=4x+20

Subtract 20 from both sides.

32-20=4x

12=4x

Divide both sides by

3=x

So, the length is

3+10=13

Therefore, the length of the rectangle is 13 cm.

7 0
3 years ago
Write an algebraic equation for this tape diagram. Then determine the value of x. You may use a calculator.
tresset_1 [31]

Answer:

3× + 16 = 22.75

3× = 22.75 - 16

<u>3x</u><u> </u>= <u>6</u><u>.</u><u>75</u>

<u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>3 3

x = 2.25

7 0
3 years ago
data set A is {30, 45, 32, 50, 33, 40, 44, 32}. Data set B is {28, 43, 30, 48, 35, 42, 46, 34}. which statement best compares th
3241004551 [841]

Answer:

the mean in set B is equal to the mean in set A (option C)

Question:

The question is incomplete as the answer choices were not given.Let's consider the following question:

Data set A is {30, 45, 32, 50, 33, 40, 44, 32}. Data set B is {28, 43, 30, 48, 35, 42, 46, 34}. which statement best compares the two data sets?

a) median for set A is equal to the median for set B

b) Range for set A is greater than range for set B 

c) The mean in set B is equal to the mean in set A

Step by step explanation:

We can describe a data set using four ways:

Center, spread, shape and unusual features.

Let's consider the center and spread.

Center: This is the median of the distribution.

Spread: This is the variation of the data set. If the range is wide, the spread is larger and If the range is small, the spread is smaller.

Rearranging the data set:

A = {30, 32, 32, 33, 40, 44, 45, 50}

B = {28, 30, 34, 35, 42, 43, 46, 48}

From the data:

The median for set A = (33+40)/2 = 73/2= 36.5

The median for set B = (35+42)/2 = 77/2= 38.5

Range = highest value - lowest value

The data ranges from 30 to 50 (range = 20) for A 

The data ranges from  28 to 48 (range = 20) for B

Mean for set A = (30+32+32+33+40+4445+50)/8 = 306/8 = 38.25

Mean for set B = (28+30+34+35+42+43+46+48)/8= 306/8 = 38.25

In both data set the mean is equal to 38.25.

Therefore the statement that best compares the two data sets is the mean in set B is equal to the mean in set A (C)

8 0
4 years ago
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