Ask question

Ask question

All categories

3 years ago

14

Hi can someone help me with the last two sets of shapes? Thank you so much! -Jenn

1 answer:

SOVA2 [1]3 years ago

8 0

You are correct there's at least one obtuse angle

You might be interested in

The speedy fast ski resort has started to keep track of number of skiers and snowboarders who bought season passes. The ratio of

pogonyaev

1250 skiers bought season passes

2500 snowboarders bought season passes

5 0

3 years ago

The number 3 was typed into cell A1 of a spreadsheet, while in cells A2-A7, formulas themselves, what would be the value display

beks73 [17]

Answer:

It's C... I just got that............................

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

a. Fill in the midpoint of each class in the column provided. b. Enter the midpoints in L1 and the frequencies in L2, and use 1-

Tresset [83]

Answer:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795

Step-by-step explanation:

Given

See attachment for class

Solving (a): Fill the midpoint of each class.

Midpoint (M) is calculated as:

$M = \frac{1}{2}(Lower + Upper)$

Where

$Lower \to$ Lower class interval

$Upper \to$ Upper class interval

So, we have:

Class 63-65:

$M = \frac{1}{2}(63 + 65) = 64$

Class 66 - 68:

$M = \frac{1}{2}(66 + 68) = 67$

When the computation is completed, the frequency distribution will be:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Solving (b): Mean and standard deviation using 1-VarStats

Using 1-VarStats, the solution is:

$\bar x = 70.2903$

$\sigma = 3.5795$

<em>See attachment for result of 1-VarStats</em>

8 0

3 years ago

Two semicircles are attached to the sides of a rectangle as shown.

melomori [17]

Answer:

$157\ in^{2}$

Step-by-step explanation:

we know that

The area of the figure is equal to the area of rectangle plus the area of two semicircles

<u>The area of rectangle is equal to</u>

$A=14*5=70\ in^{2}$

<u>The area of the small semicircle is equal to</u>

$A=\pi r^{2} /2$

$r=5/2=2.5\ in$ -----> radius is half the diameter

substitute

$A=(3.14)(2.5^{2})/2=9.8125 in^{2}$

<u>The area of the larger semicircle is equal to</u>

$A=\pi r^{2} /2$

$r=14/2=7\ in$ -----> radius is half the diameter

substitute

$A=(3.14)(7^{2})/2=76.93\ in^{2}$

The area of the figure is equal to

$70+9.8125+76.93=156.7425=157\ in^{2}$

8 0

3 years ago

Read 2 more answers

Luis and Berto sell TVs. Last month berto sold 15 more TVs. Together they sold 163. How many TVs did Luis sell?

Juli2301 [7.4K]

Luis sold 148 TVs.

Step-by-step explanation:

163-15=148

3 0

3 years ago