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vovikov84 [41]
2 years ago
6

Joshua is going to mow lawns during his summer break. The graph in the figure shows a linear relationship between the number of

weeks he will be mowing and the number of lawns he will mow. If he earns $25 for each lawn he mows, how much money will he earn after mowing lawns for 5 weeks?
Mathematics
1 answer:
inessss [21]2 years ago
7 0

Answer:

we need the graph to answer the question

Step-by-step explanation:

we wont know much he makes each week without the graph

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The scores of eighth-grade students in a math test are normally distributed with a mean of 57.5 and a standard deviation of 6.5.
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Approximately 68% of a normal distribution lies within one standard deviation of the mean, so this corresponds to students with scores between (57.5 - 6.5, 57.5 + 6.5) = (51, 64)
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The Venn diagram below provides a visual representation of the students in Mr. Griffin’s class who participate in music programs
AlekseyPX

Answer: Because the venn diagram works i donknow

Step-by-step explanation:

Go do your workk right with the diagram whatever it is

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2 years ago
An experiment was done to look at the positive arousing effects of imagery on different people. A sample of statistics lecturers
statuscvo [17]

Answer:

The Three way mixed ANOVA

Step-by-step explanation:

The Three factor repeated measures ANOVA analysis otherwise referred to as the Three way mixed ANOVA should be used for the analysis.

This is because we are comparing three independent variables in the context.

• People type (statistic lectures and students)

• Imagery type (positive, neutral and negative imagery)

• Repeated measure: Before and After images shown

• Measuring the dependent variable positive arousal as shown in the table attached to the answer.

After going through the table attached you will see that there should be performed a three factor repeated measures ANOVA analysis.

5 0
2 years ago
Traffic speed: The mean speed for a sample of cars at a certain intersection was kilometers per hour with a standard deviation o
aliina [53]

Answer:

Step-by-step explanation:

Hello!

X₁: speed of a motorcycle at a certain intersection.

n₁= 135

X[bar]₁= 33.99 km/h

S₁= 4.02 km/h

X₂: speed of a car at a certain intersection.

n₂= 42 cars

X[bar]₂= 26.56 km/h

S₂= 2.45 km/h

Assuming

X₁~N(μ₁; σ₁²)

X₂~N(μ₂; σ₂²)

and σ₁² = σ₂²

<em>A 90% confidence interval for the difference between the mean speeds, in kilometers per hour, of motorcycles and cars at this intersection is ________.</em>

The parameter of interest is μ₁-μ₂

(X[bar]₁-X[bar]₂)±t_{n_1+n_2-2} * Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }

t_{n_1+n_2-2;1-\alpha /2}= t_{175; 0.95}= 1.654

Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{134*16.1604+41*6.0025}{135+42-2} } = 3.71

[(33.99-26.56) ± 1.654 *(3.71*\sqrt{\frac{1}{135} +\frac{1}{42} })]

[6.345; 8.514]= [6.35; 8.51]km/h

<em>Construct the 98% confidence interval for the difference μ₁-μ₂ when X[bar]₁= 475.12, S₁= 43.48, X[bar]₂= 321.34, S₂= 21.60, n₁= 12, n₂= 15</em>

t_{n_1+n_2-2;1-\alpha /2}= t_{25; 0.99}= 2.485

Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{11*(43.48)^2+14*(21.60)^2}{12+15-2} } = 33.06

[(475.12-321.34) ± 2.485 *(33.06*\sqrt{\frac{1}{12} +\frac{1}{15} })]

[121.96; 185.60]

I hope this helps!

3 0
2 years ago
Endpoints of segment MN have coordinates (0, 0), (5, 1). The endpoints of segment AB have coordinates (1 1/2 , 2 1/4 ) and (−2 1
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Answer:

k=1.5

Step-by-step explanation:

3 0
2 years ago
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