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sergeinik [125]
3 years ago
14

If , which equation describes the graphed function? A. y = f(-x) − 3 B. y = -f(x) + 3 C. y = -f(x) – 3 D. y = f(-x) + 3

Mathematics
2 answers:
Elenna [48]3 years ago
7 0

Answer:

In photo below

Explanation:

In photo below

rewona [7]3 years ago
3 0

Answer:

y=-f(x)-3

Step-by-step explanation:

when f is negative it is reflected over the x-axis and -3 means to translate 3 units down

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The measures of the angles in △ABC are given by the expressions in the table.
lozanna [386]

Answer:THE CORRECT ANSWER ISSSSSS x=25

mb=65

mc=50



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How does Roosevelt use appeals to convince the
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Roosevelt appeals to the logic of the audience by noting that the interests of the people are not considered in a one-party government.

Step-by-step explanation:

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Mrs Smith is completing a mathematics problem. She knows when 6 is added to four times a number, the result is 50. What would be
Lesechka [4]

EXPLANATION:

-To formulate an equation, you must first know what data the exercise gives us to locate them correctly.

data:

-6 that must be added to a number.

-four times a number that is equal to 4x

-a result that is equal to 50

Now with these data we formulate the equation:

\begin{gathered} \text{Equation:} \\ 6+4x=50;\text{  } \end{gathered}

if we solve the equation we have:

\begin{gathered} 6\text{ }+4x=50 \\ 4x=50-6 \\ 4x=44 \\ x=\frac{44}{4} \\ x=-\frac{22}{2} \\ x=\text{ }-11 \end{gathered}

6 0
1 year ago
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
4 years ago
A school bake sale eared over $25 from selling cookies, $3 each how many cookies did they sell
MAXImum [283]
For $25 exactly is 8.3 cookies (recurring 3) but that is highly unrealistic, so over 25 dollars is 9 cookies (for 27 dollars)  at the very least.
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3 years ago
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