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alexira [117]
3 years ago
9

One

Mathematics
1 answer:
Mashcka [7]3 years ago
4 0

Someone can steal the food lol

You might be interested in
<img src="https://tex.z-dn.net/?f=%20%20%204%20%5Ctimes%20%282%20-%206%29%20%3D%20" id="TexFormula1" title=" 4 \times (2 - 6)
Brut [27]

Answer:

-16

Step-by-step explanation:

You have to start by answering the parenthesis first then continue with the question

7 0
3 years ago
Can yall solve this for me for 10 points please and thank uu
Effectus [21]

Answer:

I would say just take the big number and subtract the 41??...

4 0
3 years ago
Read 2 more answers
Select the correct answer from each drop-down menu.
ruslelena [56]

The independent variable in the relationship is the <u><em>Number of Months</em></u> and should be placed on the<em> </em><u><em>x-axis</em></u>

The dependent variable in the relationship is the <em><u>Number of Fish</u></em> and should be placed on the <u><em>y-axis</em></u>

Step-by-step explanation:

Let us explain the independent and dependent variables of a relation

  • The independent variable is the input of the relation
  • The dependent variable is the output of the relation
  • The values of dependent variables depend on the values of independent variables
  • When the relation represented graphically the independent variable placed on the x-axis and the dependent variable placed on the y-axis

The table:

→ Number of Months:  0       :  1         :  2       :  3          :  4

→ Number of Fish      : 1,024 :  1,280 :  1600 :  2,000  :  2,500

Let us check is their any constant ratio between the consecutive values of the number of fish

∵ 1,280 ÷ 1,024 = 1.25

∵ 1,600 ÷ 1.280 = 1.25

∵ 2,000 ÷ 1,600 = 1.25

∵ 2,500 ÷ 2,000 = 1.25

∴ The table represent an exponential function, where the input

   is the number of months and the output is the number of fish

∵ The input is independent variable

∴ The independent variable is the number of months

∵ The output is the dependent variable

∴ The dependent variable is the number of fish

∵ The independent variable is placed on the x-axis

∴ The number of months should be place on the x-axis

∵ The dependent variable is placed on the y-axis

∴ The number of fish should be place on the y-axis

The independent variable in the relationship is the <u><em>Number of Months</em></u> and should be placed on the<em> </em><u><em>x-axis</em></u>

The dependent variable in the relationship is the <em><u>Number of Fish</u></em> and should be placed on the <u><em>y-axis</em></u>

<u><em /></u>

Learn more:

You can learn more about the relation in brainly.com/question/10708697

#LearnwithBrainly

6 0
3 years ago
The mean weight of an adult is 69 kilograms with a variance of 121. If 31 adults are randomly selected, what is the probability
amid [387]

Answer:

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Also, important to remember that the standard deviation is the square root of the variance.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 69, \sigma = \sqrt{121} = 11, n = 31, s = \frac{11}{\sqrt{31}} = 1.97565

What is the probability that the sample mean would be greater than 70.5 kilograms?

This is 1 subtracted by the pvalue of Z when X = 70.5. So

Z = \frac{X - \mu}{\sigma}

By the Central limit theorem

Z = \frac{X - \mu}{s}

Z = \frac{70.5 - 69}{1.97565}

Z = 0.76

Z = 0.76 has a pvalue of 0.7764

1 - 0.7764 = 0.2236

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

8 0
3 years ago
Vector component of ū and w are as
Helen [10]

Answer:

r = ( -14.4 , 6.47 )

Step-by-step explanation:

in the photo you will find all the steps

7 0
2 years ago
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