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ivann1987 [24]
4 years ago
15

Help me with this problem​

Mathematics
1 answer:
Dima020 [189]4 years ago
4 0

Answer:

a. Ryan 70mph. Janelle 65mph.

b. 4 hours and 48 minutes.

Step-by-step explanation:

a. Ryan 245/3.5=70 mph. Janelle 260/4=65 mph.

b. If Janelle drives 575 miles with the speed of 65 mph, in total it would take her 575/65=8.8 hours, which is 4.8 hours more than 4 hours. (0.8 hours is 48 minutes).

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I need to know how to do the whole thing and understand it.
patriot [66]

We are given the data on the number of candies handed by neighborhood A and neighborhood B.

Let us first find the mean and variance of each neighborhood.

Mean:

\bar{x}_A=\frac{\sum x}{N_1}=\frac{12}{6}=2\bar{x}_B=\frac{\sum x}{N_2}=\frac{20}{6}=3.33

Variance:

s_A^2=\frac{\sum x^2}{N_1}-\bar{x}_A^2=\frac{28}{6}-2^2=0.667s_B^2=\frac{\sum x^2}{N_2}-\bar{x}_B^2=\frac{80}{6}-3.33^2=2.244

A. Null hypothesis:

The null hypothesis is that there is no difference in the mean number of candies handed out by neighborhoods A and B.

H_0:\;\mu_A=\mu_B

Research hypothesis:

The research hypothesis is that the mean number of candies handed out by neighborhood A is more than neighborhood B.

H_a:\;\mu_A>\mu_B

Test statistic (t):

The test statistic of a two-sample t-test is given by

t=\frac{\bar{x}_A-\bar{x}_B}{s_p}

Where sp is the pooled standard deviation given by

\begin{gathered} s_p=\sqrt{\frac{N_1s_1^2+N_2s_2^2}{N_1+N_2-2}(\frac{N_1+N_2}{N_1\cdot N_2}}) \\ s_p=\sqrt{\frac{6\cdot0.667+6\cdot2.244}{6+6-2}(\frac{6+6}{6\cdot6})} \\ s_p=0.763 \end{gathered}t=\frac{2-3.33}{0.763}=-1.74

So, the test statistic is -1.74

Critical t:

Degree of freedom = N1 + N2 - 2 = 6+6-2 = 10

Level of significance = 0.05

The right-tailed critical value for α = 0.05 and df = 10 is found to be 1.81

Critical t = 1.81

We will reject the null hypothesis because the calculated t-value is less than the critical value.

Interpretation:

This means that we do not have enough evidence to conclude that neighborhood A gives out more candies than neighborhood B.

6 0
1 year ago
22. A liver cell is about 20 um in diameter. How many liver cells would it take to form an end to end chain 10 in long?
34kurt

The number of liver cells it would take to form an end to end chain 10 in long is 12,700 liver cells

<h3>Calculating quantity</h3>

From the question, we are to determine how many liver cells it would take to form an end to end chain 10 in long

First, we will convert 10 in to meters

1 inch = 0.0254

∴ 10 inches = 0.254 m

From the giving information,

A liver cell is about 20 μm in diameter

∴ The number of liver cells it would take to form an end to end chain 10 in long = 0.254 m / 20 μm

= 0.254 m / (20 ×10⁻⁶) m (NOTE: 1μm = 10⁻⁶m)

= 12700

Hence, the number of liver cells it would take to form an end to end chain 10 in long is 12,700 liver cells

Learn more on Calculating quantity here: brainly.com/question/2021001

#SPJ1

8 0
2 years ago
Can someone help me?
wlad13 [49]
V= pi r^2h
180= pi (5)^2h
180=78.5h
Divide by 78.5
h= 2.29
h= 2.3
Sorry could not type pi symbol
5 0
4 years ago
Find all the common factors of 42 and 21 and identify the greatest common factor​
Anuta_ua [19.1K]

Answer:

The factors of 21 are: 1, 3, 7, 21

The factors of 42 are: 1, 2, 3, 6, 7, 14, 21, 42

21 is the Greatest common factor

Step-by-step explanation:

5 0
3 years ago
Help me on the first question
Tresset [83]
34.7 - (12.07 + 4.9) = 
34.7 - 16.97 =
17.73 <===
7 0
3 years ago
Read 2 more answers
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