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3 years ago

Can you please show work

Mathematics

2 answers:

stepan [7]3 years ago

6 0

Answer:

A) 3.44 ft²

Step-by-step explanation:

before cutouts, the area was 4 x 4 = 16 ft²

the circles have a diameter of 2 ft each, and a radius of 1 ft

each circle has an area pf (3.14)(1²) = 3.14 ft²

3.14 x 4 = 12.56 ft²

16 - 12.56 = 3.44 ft² left

aleksley [76]3 years ago

5 0

Answer:area of 4 equal circle having

Diameter =2ft

Therefore radius =r=2/2=1ft

Area of 4 equal circle =4(πr²) =4π×1²=4πft²

Area of square=length²=4²=16

Area of shaded region =area of square-

area of 4 equal circle

=16-4π=3.43362938564ft²

Step-by-step explanation:

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For questions 4–5, simplify the expression.

Alex17521 [72]

1. $\mathrm{Group\:like\:terms} \ \textgreater \ 24h-10h+3d+8d$

$\mathrm{Add\:similar\:elements:}\:24h-10h=14h \ \textgreater \ 14h+3d+8d$

$\mathrm{Add\:similar\:elements:}\:3d+8d=11d \ \textgreater \ 14h+11d$

2. $\mathrm{Group\:like\:terms} \ \textgreater \ 12s+4s-2+4$

$\mathrm{Add\:similar\:elements:}\:12s+4s=16s \ \textgreater \ 16s-2+4$

$\mathrm{Add/Subtract\:the\:numbers:}\:-2+4=2 \ \textgreater \ 16s+2$

Hope this helps!

4 0

3 years ago

Which expression(s) represent negative products? Check all that apply. 7.71(–0.02) 3.4(0.5) 0.23(–1.1) –2.5(–9.2) –5.09(8.1)

andriy [413]

Answer: 1 3 5

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

The ages of a random sample of five university professors are 39, 54, 61, 72, and 59. Using this information, find a 99% confide

kondor19780726 [428]

Answer:

99% confidence interval for the population standard deviation = (74.97 , 635.20).

Step-by-step explanation:

We are given that the ages of a random sample of five university professors are 39, 54, 61, 72 and 59. Also, it is provided that the ages of university professors are normally distributed.

So, firstly the pivotal quantity for 99% confidence interval for the population standard deviation is given by;

P.Q. = $\frac{(n-1)s^{2} }{\sigma^{2} }$ ~ $\chi^{2} __n_-_1$

where, s = sample standard deviation

$\sigma$ = population standard deviation

n = sample of university professors = 5

Also, $s^{2}$ = $\frac{\sum (X-\bar X)^{2} }{n-1}$ = 144.5

So, 99% confidence interval for population standard deviation, $\sigma$ is;

P(0.2070 < $\chi^{2} __5_-_1$ < 14.86) = 0.99 {As the table of $\chi^{2}$ at 4 degree of freedom

gives critical values of 0.2070 & 14.86}

P(0.2070 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 14.86) = 0.99

P( $\frac{ 0.2070}{(n-1)s^{2} }$ < $\frac{1 }{\sigma^{2} }$ < $\frac{ 14.86}{(n-1)s^{2} }$ ) = 0.99

P( $\frac{ (n-1)s^{2}}{14.86 }$ < $\sigma^{2}$ < $\frac{ (n-1)s^{2}}{0.2070 }$ ) = 0.99

99% confidence interval for $\sigma^{2}$ = ( $\frac{ (n-1)s^{2}}{14.86 }$ , $\frac{ (n-1)s^{2}}{0.2070 }$ )

= ( $\frac{ (5-1) \times 144.5^{2}}{14.86 }$ , $\frac{ (5-1) \times 144.5^{2}}{0.2070 }$ )

= (5620.525 , 403483.092)

99% confidence interval for $\sigma$ = ( $\sqrt{5620.525}$ , $\sqrt{403483.092}$ )

= (74.97 , 635.20)

Therefore, 99% confidence interval for the population standard deviation of the ages of all professors at the university is (74.97 , 635.20).

8 0

4 years ago

How many milligrams of medication are in 75 mL of a 2% solution?<br><br> Help

deff fn [24]

The answer is 23 I took the test

4 0

3 years ago

Lee's family had a garden in the shape of a right triangle. the total area of the garden is 16 square feet. if they expand the g

lana [24]

Area of right-angled triangle is given by;

Area, A = 1/2 *b*h, Where b=base, h=height

Therefore,

A1 = 1/2bh = 16 in^2
A2 = 1/2 (2b)(2h) = 2bh

Ratio of increase = A2/A1 = {2bh}/(1/2bh} = 4 (the area is increased 4 times)
The,
A2 = 4*16 = 64 in^2
Therefore,
The area is increased by (64-16) = 48 in^2

5 0

3 years ago

Can you please show work​

Can you please show work