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3 years ago

The table lists the locations of four underwater cameras relative to sea level which is the highest

Mathematics

1 answer:

atroni [7]3 years ago

6 0

Answer: Turtle Camera

Step-by-step explanation:

The Turtle is the closest to 0, at around 5. Everything else is higher.

Hope that helped,

-sirswagger21

P.S. Please mark this as brainliest answer, it would help a lot!

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the movie shop sells posters for $7 dvd for $15 and cd for $9 write an expression to represent the total cost for 2 posters d dv

Nat2105 [25]

Try (7x2)+d+c
I hope this helped?
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3 years ago

A rectangular room has a perimeter of 70 m and is 20m long. write and equation to represent the area

REY [17]

70m+20m+90m could be one

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3 years ago

Read 2 more answers

(4,7), (x,19); slope = 3 the pair points lies on the same line with the given slope. Find x

Andrew [12]

So we are given two points, say P1(4,7), P2(x,19).

Slope is given by
m=3=(y2-y1)/(x2-x1)=(19-7)/(x-4)
solve for x
3=(19-7)/(x-4)
cross multiply
3(x-4)=12
3x-12=12
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x=8

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4 years ago

I-Ready Surface area of composed figures what is the surface area of this design

PilotLPTM [1.2K]

Answer: 512

Step-by-step explanation:

Hello, I hope the answer is correct if not let me know and i will work step by step on it with you. Your welcome and Happy to help.

8 0

3 years ago

When the author visited Dublin, Ireland (home of Guinness Brewery employee William Gosset, who first developed the t distributio

disa [49]

Answer:

The Decision Rule

Fail to reject the null hypothesis

The conclusion

There is no sufficient evidence to support the claim that the mean age of the cars is greater than that of taxi

Step-by-step explanation:

From the question we are told that

The data is

Car Ages 4 0 8 11 14 3 4 4 3 5 8 3 3 7 4 6 6 1 8 2 15 11 4 1 6 1 8

Taxi Ages 8 8 0 3 8 4 3 3 6 11 7 7 6 9 5 10 8 4 3 4

The level of significance $\alpha = 0.05$

Generally the null hypothesis is $H_o : \mu_1 - \mu_2 = 0$

the alternative hypothesis is $H_a : \mu_1 - \mu_2 > 0$

Generally the sample mean for the age of cars is mathematically represented as

$\= x_1 = \frac{\sum x_i }{n}$

=> $\= x_1 = \frac{4+ 0+ 8 +11 + \cdots + 8
}{27}$

=> $\= x_1 = 5.56$

Generally the standard deviation of age of cars

$\sigma _1 = \sqrt{\frac{\sum (x_i - \= x)^2}{n_1} }$

=> $\sigma _1 = \sqrt{\frac{(4 - 5.56)^2 + (0 - 5.56)^2+ (8 - 5.56)^2 + \cdots + 8}{ 27} }$

=> $\sigma _1 = 3.88$

Generally the sample mean for the age of taxi is mathematically represented as

$\= x_2 = \frac{\sum x_i }{n}$

=> $\= x_2 = \frac{8 +8 +0 + \cdots + 4
}{20}$

=> $\= x_2 = 5.85$

Generally the standard deviation of age of taxi

$\sigma _2 = \sqrt{\frac{\sum (x_i - \= x)^2}{n_1} }$

=> $\sigma _2 = \sqrt{\frac{(8 - 5.85)^2 + (8 - 5.85)^2+ (0 - 5.85)^2 + \cdots + 8}{ 20} }$

=> $\sigma _2 = 2.83$

Generally the test statistics is mathematically represented as

$t = \frac{(\= x_ 1 - \= x_2 ) - 0}{\sqrt{\frac{\sigma^2_1}{n_1} + \frac{\sigma^2_2}{n_2} } }$

=> $t = \frac{(5.56 - 5.85 ) - 0}{\sqrt{\frac{(3.88)^2}{27} + \frac{(2.83)}{20} }}$

=> $t = -0.30$

Generally the degree of freedom is mathematically represented as

$df = n_1 + n_2 -2$

$df = 27 + 20 -2$

$df = 45$

From the t distribution table the $P(t > t )$ at the obtained degree of freedom = 45 is

$P(t > -0.30 ) = 0.61722067$

So the p-value is

$p-value = P(t > T) = 0.61722067$

From the obtained values we see that the p-value > $\alpha$ hence we fail to reject the null hypothesis

Hence the there is no sufficient evidence to support the claim that the mean age of the cars is greater than that of taxi

5 0

3 years ago