All categories

3 years ago

PLEASE HELP ASAP.. WILL FAN AND MEDAL

Mathematics

2 answers:

oksian1 [2.3K]3 years ago

6 0

Answer: 17.07 centimeters, because height = Square root of the difference of the squares of 18 and 5.7.

Step-by-step explanation:

Given: The slant height of the cone (l)= 18 cm

The radius of the cone (r) =5.7 cm

Let h be the height of the cone .

We know that the above dimensions make aright triangle in a cone, such that by Pythagoras theorem

$l^2=h^2-r^2\\\Rightarrow\ h^2=l^2-r^2\\\Rightarrow\ h=\sqrt{l^2-r^2}\\\Rightarrow\ h=\sqrt{18^2-5.7^2}$

Thus, height = Square root of the difference of the squares of 18 and 5.7

and,

$\Rightarrow\ h=\sqrt{324-32.49}\\\Rightarrow\ h=\sqrt{291.51}\\\Rightarrow\ h=17.0736639302\approx17.07\ cm$

Komok [63]3 years ago

4 0

C. 18.88 centimeters. Because with a tilted radius of 5.7. It ranges up. So it would be a higher number SLIGHTLY.
Hope this helps.

You might be interested in

Solve the right triangle

hoa [83]

Answer:

is the right one

Step-by-step explanation:

8 0

3 years ago

A person's hourly pay rate, r, is found by dividing the total amount

marishachu [46]

Translating the words into formula, we have

$r = \dfrac{p}{w}$

In fact, we have that the pay rate is the total payment divided by the number of hours worked.

We can solve this formula for w in two steps: starting from the formula above, multiply both sides by w:

$rw = p$

And then divide both sides by r:

$w = \dfrac{p}{r}$

4 0

3 years ago

Help me plz.................

EastWind [94]

Answer:

1/8 cup

Step-by-step explanation:

x = 7/8 - 1/4 - 1/2

x = 7/8 - 2/8 - 4/8

x = 1/8

5 0

3 years ago

Read 2 more answers

Dannette and Alphonso work for a computer repair company. They must include the time it takes to complete each repair in their r

olga55 [171]

Answer:

(a):

Dannette Alphonso

$\bar x_D = 4.33$ $\bar x_A = 5.17$

$M_D = 2.5$ $M_A = 5$

$\sigma_D = 3.350$ $\sigma_A = 1.951$

$IQR_D = 7$ $IQR_A = 1.5$

(b):

Measure of center: Median

Measure of spread: Interquartile range

(c):

There are no outliers in Dannette's dataset

There are outliers in Alphonso's dataset

Step-by-step explanation:

Given

See attachment for the appropriate data presentation

Solving (a): Mean, Median, Standard deviation and IQR of each

From the attached plots, we have:

IQR_A = 1.5 ---- Dannette

$A = \{3,4,4,4,4,5,5,5,5,6,6,11\}$ ---- Alphonso

n = 12 --- number of dataset

Mean

The mean is calculated

$\bar x = \frac{\sum x}{n}$

So, we have:

$\bar x_D = \frac{1+1+1+1+2+2+3+7+8+8+9+9}{12}$

$\bar x_D = \frac{52}{12}$

$\bar x_D = 4.33$ --- Dannette

$\bar x_A = \frac{3+4+4+4+4+5+5+5+5+6+6+11}{12}$

$\bar x_A = \frac{62}{12}$

$\bar x_A = 5.17$ --- Alphonso

Median

The median is calculated as:

$M = \frac{n + 1}{2}th$

$M = \frac{12 + 1}{2}th$

$M = \frac{13}{2}th$

$M = 6.5th$

This implies that the median is the mean of the 6th and the 7th item.

So, we have:

$M_D = \frac{2+3}{2}$

$M_D = \frac{5}{2}$

$M_D = 2.5$ ---- Dannette

$M_A = \frac{5+5}{2}$

$M_A = \frac{10}{2}$

$M_A = 5$ ---- Alphonso

Standard Deviation

This is calculated as:

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n}}$

So, we have:

$\sigma_D = \sqrt{\frac{(1 - 4.33)^2 +.............+(9- 4.33)^2}{12}}$

$\sigma_D = \sqrt{\frac{134.6668}{12}}$

$\sigma_D = 3.350$ ---- Dannette

$\sigma_A = \sqrt{\frac{(3-5.17)^2+............+(11-5.17)^2}{12}}$

$\sigma_A = \sqrt{\frac{45.6668}{12}}$

$\sigma_A = 1.951$ --- Alphonso

The Interquartile Range (IQR)

This is calculated as:

$IQR =Q_3 - Q_1$

Where

$Q_3 \to$ Upper Quartile and $Q_1 \to$ Lower Quartile

$Q_3$ is calculated as:

$Q_3 = \frac{3}{4}*({n + 1})th$

$Q_3 = \frac{3}{4}*(12 + 1})th$

$Q_3 = \frac{3}{4}*13th$

$Q_3 = 9.75th$

This means that $Q_3$ is the mean of the 9th and 7th item. So, we have:

$Q_3 = \frac{1}{2} * (8+8) = \frac{1}{2} * 16$ $Q_3 = \frac{1}{2} * (5+6) = \frac{1}{2} * 11$

$Q_3 = 8$ ---- Dannette $Q_3 = 5.5$ --- Alphonso

$Q_1$ is calculated as:

$Q_1 = \frac{1}{4}*({n + 1})th$

$Q_1 = \frac{1}{4}*({12 + 1})th$

$Q_1 = \frac{1}{4}*13th$

$Q_1 = 3.25th$

This means that $Q_1$ is the mean of the 3rd and 4th item. So, we have:

$Q_1 = \frac{1}{2}(1+1) = \frac{1}{2} * 2$ $Q_1 = \frac{1}{2}(4+4) = \frac{1}{2} * 8$

$Q_1 = 1$ --- Dannette $Q_1 = 4$ ---- Alphonso

So, the IQR is:

$IQR = Q_3 - Q_1$

$IQR_D = 8 - 1$ $IQR_A = 5.5 - 4$

$IQR_D = 7$ --- Dannette $IQR_A = 1.5$ --- Alphonso

Solving (b): The measures to compare

Measure of center

By observation, we can see that there are outliers is the plot of Alphonso (because 11 is far from the other dataset) while there are no outliers in Dannette plot (as all data are close).

Since, the above is the case; we simply compare the median of both because it is not affected by outliers

Measure of spread

Compare the interquartile range of both, as it is arguably the best measure of spread, because it is also not affected by outliers.

Solving (c): Check for outlier

To check for outlier, we make use of the following formulas:

$Lower =Q_1 - 1.5 * IQR$

$Upper =Q_3 + 1.5 * IQR$

For Dannette:

$Lower = 1 - 1.5 * 7 = -9.5$

$Upper = 8 + 1.5 * 7 = 18.5$

Since, the dataset are all positive, we change the lower outlier to 0.

So, the valid data range are:

$Valid = 0 \to 18.5$

From the question, the range of Dannette's dataset is: 1 to 9. Hence, there are no outliers in Dannette's dataset

For Alphonso:

$Lower = 4 - 1.5 * 1.5 =1.75$

$Upper = 5.5 + 1.5 * 1.5 =7.75$

So, the valid data range are:

$Valid = 1.75\to 7.75$

From the question, the range of Alphonso's dataset is: 3 to 11. Hence, there are outliers in Alphonso's dataset

4 0

3 years ago

Show your work and the answer please

Fofino [41]

Answer:

-0.375

Step-by-step explanation:

Writing -3/8 as a decimal using division method:

To convert any fraction to decimal form, we just need to divide its numerator by denominator.

Here, the fraction is -3/8 which means we need to perform 3 ÷ 8.

This gives the answer as -0.375. So, -3/8 as a decimal is -0.375

7 0

2 years ago