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Bess [88]
3 years ago
11

One box holds 16 crayons. Another box hold 24 crayons. Write a sentence using words that compares the crayons in the two boxes

Mathematics
2 answers:
goldenfox [79]3 years ago
6 0
Box 1 has 8 less than the box with 24 in it.
levacccp [35]3 years ago
6 0
24 crayons is are greater than 16 crayons
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Luna at $50 when she got to the carnival after riding 12 rides she had twenty-six dollars left what was the price of which equat
Finger [1]

\frac{50 - 26}{12}  = x
\frac{24}{12}  = x
2 = x
8 0
3 years ago
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Select strong association, weak association, or moderate association to correctly classify the correlation coefficient.
11Alexandr11 [23.1K]
Answers:


0.45 is a moderate association

0.95 and -0.8 are both strong association

0.10 is weak association


Explanation:


This is the interpreation of the correltaion coefficient:

1) The correlaion coefficient assesses the relationship between two variables in a scatter plot.


2) If the sign of the correlation coefficient is positive means that the two variables trend to grow or decrease in the same sense. This is an uphill line or curve:  if variable X grows, variable Y grows, and if variable X decreases variable Y grows.

If the sign of the correlation coefficient is negative means that the two variables go in opposite direction. This is a downhill line or curve.


3) A correlation coefficient of +1 or -1 is a perfect association. The two variables are totally associated.


4) A correlation coefficient less that +1 but greater than 0.7 is a strong association. The same with a coefficite between - 0.7 and -1.


5) A correlation coefficient arroun +0.5 or -0.5 is a moderate association.


6) A correlation coefficient of 0 is a nill association.


7) A correlation coeffiicient between 0 and 0.3 is a weak association. The same when the correlation coefficient is between -3 and 0.
4 0
3 years ago
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5) Contains (3,-1) and (6,- 13)<br> Help please
zimovet [89]

Answer:

The slope of the line is -12/3

Step-by-step explanation:

remember, rise over run

My I have brainliest please? :)

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3 years ago
How do I give a Brainliest to somebody?
Dominik [7]

Answer:

there have to be at least two answers, then you have the option to mark brainliest by pressing the crown

Step-by-step explanation:

Hope this helps!

6 0
2 years ago
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Consider a population of Pacific tree frogs (Pseudacris regilla). In this population of frogs, a single locus controls the produ
zloy xaker [14]

Answer:

Step-by-step explanation:

Hello!

Use these data to calculate the chi-square statistic. Report your calculated chi-square statistic to two decimal places.

In this example, the study variable is X: genotype of a single locus two allele gene that codes the mating call of tree frogs. Categorized: S₁S₁, S₁S₂, and S₂S₂.

Usually, when you have a sample of observed genotypes of a gene of interest, the goal is to check if these genotypes follow a theoretical model such as the frequency models proposed by Mendel.

In Mendelian genetics, when two heterozygous individuals (Aa, Aa) from the F1 generation are crossed, you expect the next generation (F2) to show the genotypic ratio 1:2:1 ⇒ This model means that out of 4 descendants 1 will have the genotype AA, 2 will have the genotype Aa and 1 will have the genotype aa. Since there is no theoretical model specified I'll use the mendelian ratio 1:2:1.

If this theory applies to the population then we'll expect that the proportion of the first genotype is P(S₁S₁)= 0.25, the second P(S₁S₂)= 0.5 and the third P(S₂S₂)=0.25

To test if the observed genotypes follow the theoretical model you have to apply a Chi-Square Goodness of Fit Test.

For this test the hypotheses are:

H₀:P(S₁S₁)= 0.25; P(S₁S₂)= 0.5; P(S₂S₂)=0.25

H₁: The data is not consistent with the specified distribution.

α: 0.05

Statistic: X^2= sum(\frac{(o_i-e_i)^2}{e_i} )~X^2_{k-1}

Where:

oi: Observed frequency for the i- category

ei: Expected frequency for the i-category

k= number of categories

The rejection region for this test is one-tailed to the right. This is so because if the observed and expected values are too different (the chi-square value will be high) this will mean that the population doesn't follow the theoretical model and thus reject the null hypothesis. If the differences between what's observed and what's expected are small, this will mean that the population follows the theoretical model (and you'll obtain a small chi-square value) and you will not reject the null hypothesis.

The critical value is:

X^2_{k-1; 1 - \alpha }= X^2_{3;0.95}= 7.815

If the statistic is at least 7.815, the decision is to reject the null hypothesis.

If the statistic is less than 7.815, the decision is to not reject the null hypothesis.

Step 1 is to obtain the expected frequencies for each category:

e_i= n*P_i

e_{S_1S_1}= n * P(S_1S_1)= 1341*0.25= 335.25

e_{S_1S_2}= n * P(S_1S_2)= 1341 * 0.5= 670.5

e_{S_2S_2}= n * P(S_2S_2)= 1341*0.25= 335.25

Note: the summatory of the expected frequencies is equal to the total of observations ∑ei= n. If you have doubts about your calculations, you can add them to check it: ∑ei= 335.25 + 670.5 + 335.25= 1341

X^2= (\frac{(987-335.25)^2}{335.25} )+(\frac{(333-670.5)^2}{670.5} )+(\frac{(21-335.25)^2}{335.25} ) = 1731.50

Decision: Reject null hypothesis.

I hope this helps!

3 0
3 years ago
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