Find the degree of the monomial. 3a4b3
1 answer:
You might be interested in
Given :
400 reduced by 3 times my age is 241
To Find :
My age .
Solution :
Let, age is x .
So , expressing given condition mathematically , we get :
Therefore, the age is 53 years.
Hence, this is the required solution.
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: ~
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:
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:
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
Decision: Reject null hypothesis.
I hope this helps!
Answer:
22.96% probability that a hotel room costs between $250.00 and $285.00
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
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.
In this problem, we have that:
What is the probability that a hotel room costs between $250.00 and $285.00?
This is the pvalue of Z when X = 285 subtracted by the pvalue of Z when X = 250. So
X = 285
has a pvalue of 0.7734
X = 250
has a pvalue of 0.5438
0.7734 - 0.5438 = 0.2296
22.96% probability that a hotel room costs between $250.00 and $285.00