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Mkey [24]
3 years ago
10

Help! Help!

Mathematics
1 answer:
Nina [5.8K]3 years ago
8 0

Answer:

1. 50%

2.100%

3.0%

4.100%

Step-by-step explanation:

Use https://www.mathsisfun.com/data/standard-deviation-calculator.html for later references.

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One side of a roof is in the shape of a trapezoid (similar to the one on Pizza Hut). The top row requires 10 shingles, the secon
larisa [96]
<span> It would depend on the number of rows?

This is an arithmetic sequence and the sum of n terms is --> n/2(2a+(n-1)d), where a is the first term and d is the common difference. In this case a = 10, d =6

n/2(20+6(n-1)) = 3n^2 + 7n = ANSWER

n = number of rows

</span>
6 0
3 years ago
Listed below are annual data for various years. The data are weights​ (metric tons) of imported lemons and car crash fatality ra
anyanavicka [17]

Answer:

Because the​ P-value is _(<u>0.02)  less </u> than the significance level 0.05​, there <u> is </u> sufficient evidence to support the claim that there is a linear correlation between lemon imports and crash fatality rates for a significance level of α=0.05.

C. The results suggest that imported lemons cause car fatalities.

Step-by-step explanation:

Hello!

The study variables are:

X₁: Weight of imported lemons.

X₂: Car crash fatality rate.

The objective is to test if the imported lemons affect the occurrence of car fatalities. To do so you are asked to use a linear correlation test.

I've made a Scatterplot with the given data, it is attached to the answer.

To be able to use the parametric linear correlation you can use the parametric test (Person) or the nonparametric test Spearman. For Person, you need your variables to have a bivariate normal distribution. Since one of the variables is a discrete variable (ratio of car crashes) and the sample is way too small to make an approximation to a normal distribution, the best test to use is Spearman's rank correlation.

This correlation coefficient (rs) takes values from -1 to 1

If rs = -1 this means that there is a negative correlation between the variables

If rs= 1 this means there is a positive correlation between the variables

If rs =0 then there is no correlation between the variables.

The hypothesis is:

H₀: There is no linear association between X₁ and X₂

H₁: There is a linear association between X₁ and X₂

α: 0.05

To calculate the Spearman's correlation coefficient you have to assign ranks to the observed values of each variable, from the smallest to the highest). Then you have to calculate the difference (d)between the ranks and the square of that difference (d²). (see attachment)

The formula for the correlation coefficient is:

rs= 1 - \frac{6* (sum of d^2)}{(n-1)n(n+1)}

rs= 1 - \frac{6* (40)}{4*5*6}

rs= -1

For this value of the correlation coefficient, the p-value is 0.02

Since the p-value (0.02) is less than the significance level (0.05) the decision is to reject the null hypothesis. In other words, there is a linear correlation between the imported lemons and the car crash fatality ration, this means that the modification in the lemon import will affect the car crash fatality ratio.

Note: the correlation coefficient is negative, so you could say that there is a correlation between the variables and this is negative (meaning that when the lemon import increases, the car crash fatality ratio decreases)

I hope it helps!

4 0
3 years ago
What is the quotient?<br><br> -4/5 *2
kogti [31]

Answer:

-8/5

Step-by-step explanation:

convert the expression:-4/5 *2

multiply the fractions: -4/5 *2/1 (multiply across)

the you will get -8/5

4 0
3 years ago
A. A 5in radius ball is in a
abruzzese [7]

Answer:

52.33%

Step-by-step explanation:

Since, all six sides of the box are touching the ball.

So, all the three dimensions of the box are equal.

Hence, it is a cubical box.

Side length of the box (l)

= diameter of the ball

= 2*radius of the ball

= 2*5

= 10 In.

V_{ball} = \frac{4}{3} \pi r^3

V_{ball} = \frac{4}{3} \times 3.14(5)^3

V_{ball} = \frac{4}{3} \times 3.14\times 125

V_{ball} = \frac{1,570}{3}\: In^3

V_{box} =l^3

V_{box} =(10)^3

V_{box} =1000\: In^3

Percentage of the space in the box occupied by the ball

= \frac{V_{ball}}{V_{box} } \times 100

= \frac{\frac{1,570}{3}}{1000} \times 100

= \frac{1570}{3000} \times 100

= \frac{157}{3}

= 52.33\%

5 0
3 years ago
Use partial quotient 28 divided by 514
Ann [662]
The result is 18 r 10
5 0
3 years ago
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