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Semmy [17]
3 years ago
9

there are 20 oranges in a box .if 5 oranges are rotten, calculate what percentage of the oranges are good need all calculations

please ​
Mathematics
2 answers:
telo118 [61]3 years ago
5 0

20 total oranges

5 are bad.

Subtract the bad from the total to find the number of good ones:

20-5 = 15

There are 15 good oranges out of 20 total.

Divide good by total then multiply by 100 to get the percentage:

15/20 = 0.75

0.75 x 100 = 75%

Answer: 75% are good

puteri [66]3 years ago
5 0

Answer:

75% are good

Step-by-step explanation:

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I would say 5 kg because you get more kg with a little bit more money added

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2 years ago
Can somebody help me ? please .. if you answer you’ll get 20 points & brainlyiest if correct! ( sorry i might have spelled b
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Step-by-step explanation:

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3 years ago
The equation of a linear function in point-slope form is y-y1=m(x-x1). Harold correctly wrote the equation y=3(x-7) using a poin
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Answer:

(7, 0).

Step-by-step explanation:

We have   y =  3(x - 7)

y = 3x - 21

Point-slope form is

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y  -  y1 = 3x - 3x1

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5 0
3 years ago
A simple random sample of 28 Lego sets is obtained and the number of pieces in each set was counted.The sample has a standard de
Karo-lina-s [1.5K]

Answer:

Step-by-step explanation:

Given that:

A simple random sample n = 28

sample standard deviation S = 12.65

standard deviation \sigma = 11.53

Level of significance ∝ = 0.05

The objective is to test the claim that the number of pieces in a set has a standard deviation different from 11.53.

The null hypothesis and the alternative hypothesis can be computed as follows:

Null hypothesis:

H_0: \sigma^2 = \sigma_0^2

Alternative hypothesis:

H_1: \sigma^2 \neq \sigma_0^2

The test statistics can be determined by using the following formula in order to test if the claim is statistically significant or not.

X_0^2 = \dfrac{(n-1)S^2}{\sigma_0^2}

X_0^2 = \dfrac{(28-1)(12.65)^2}{(11.53)^2}

X_0^2 = \dfrac{(27)(160.0225)}{132.9409}

X_0^2 = \dfrac{4320.6075}{132.9409}

X_0^2 = 32.5002125

X^2_{1- \alpha/2 , df} = X^2_{1- 0.05/2 , n-1}

X^2_{1- \alpha/2 , df} = X^2_{1- 0.025 , 28-1}

From the chi-square probabilities table at 0.975 and degree of freedom 27;

X^2_{0.975 , 27} = 14.573

X^2_{\alpha/2 , df} = X^2_{ 0.05/2 , n-1}

X^2_{\alpha/2 , df} = X^2_{0.025 , 28-1}

From the chi-square probabilities table at 0.975 and degree of freedom 27;

X^2_{0.025 , 27}= 43.195

Decision Rule: To reject the null hypothesis if X^2_0  \ >  \ X^2_{\alpha/2 , df}  \ \  \ or \ \ \   X^2_0 \  < \  X^2_{1- \alpha/2 , df} ; otherwise , do not reject the null hypothesis:

The rejection region is X^2_0  \ >  43.195 \ \  \ or \ \ \   X^2_0 \  < \  14.573

Conclusion:

We fail to reject the null hypothesis since  test statistic value 32.5002125  lies  between 14.573 and 43.195.

6 0
3 years ago
How do you know when there are no solutions to a system of equations?
Allushta [10]
No solution
========
The lines are parallel
eg
y = 2x + 3; y = 2x + 5

Infinite solutions
============
When one equation reduces to the second one.
2y = 10x + 6
y = 5x + 3 When you divide the first one by 2, you get the second one.
4 0
3 years ago
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