HELP ME OUT PLEASE!!!

The probability that a person who has made a reservation for a trip on a twelve-vehicle ferry will actually arrive and make the

skad [1K]

Answer:

D

Step-by-step explanation:

5 0

3 years ago

Thank you everyone for helping me this year to get good grades I am giving points.

Dmitriy789 [7]

Yayyyy I’m so happy you got good grades

6 0

2 years ago

Late work 4 physics

Oliga [24]

Answer:

Step-by-step explanation:

problem 4

Velocity is +2.0m/s and constant acceleration is -0.5m/s^2

For every second the velocity drop by 0.5 m/s, so after 2 seconds the velocity is 1 m/s.

problem 5

Accelerate from rest to 28m/s means go from 0 to 28m/s so with

a constant acceleration of 5.5 m/s^2 it will take 5.09 second

7 0

3 years ago

The heights of dogs, in inches, in a city are normally distributed with a population standard deviation of 3 inches and an unkno

Nitella [24]

Answer: 1.2703

Step-by-step explanation:

Given : Sample size : n= 17, which is a small sample (n<30), so we use t-test.

Significance level : $\alpha:1-0.90=0.1$

Then , Critical value : $t_{n-1,\alpha/2}=t_{16, 0.05}\pm1.745884$

Standard deviation: $s=3\text{ inches}$

The formula to find the margin of error : -

$E=t_{n-1,\alpha/2}\dfrac{s}{\sqrt{n}}\\\\\Rightarrow\ E=(1.745884)\dfrac{3}{\sqrt{17}}\\\\\Rightarrow\ E=01.27031720154\approx1.2703$

Hence, the error bound (EBM) of the confidence interval with a 90% confidence level.=1.2703

6 0

2 years 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

3 years ago