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3 years ago

If you can't see it the question is use benchmarks to estimate decimals:

Mathematics

2 answers:

adoni [48]3 years ago

8 0

So do you want me to estimate with benchmarks or figur the whole thing

Svetllana [295]3 years ago

6 0

10.29 is the answer if you were asking that

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Find the surface area of the prism or pyramid. Round to the nearest hundredth if necessary.

son4ous [18]

Answer: 345.4

Step-by-step explanation:

SA=2πr^2+2πrh

SA=2(3.14)(5)^2+2(3.14)(5)(6)

SA=2(3.14)(25)+2(3.14)(30)

SA=2(78.5)+2(94.2)

SA=157+188.4

SA=345.4

4 0

3 years ago

I NEED AN ANSWER NOW ASAP

babunello [35]

Answer:

Apparently i cant see the photo im am so sorry but do need pints.

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Use Heron's Area Formula to find the area of the triangle. (Round your answer to two decimal places.)

aniked [119]

Answer:

1549.24

Step-by-step explanation:

Heron's Area formula:

A = $\sqrt{s(s - a)(s - b)(s - c)}$ , where s is semi-perimeter
s = 1/2(a + b + c)

We are given two sides and the included angle.

Use the law of cosines to find the missing side:

a² = b² + c² - 2bccos A
a² = 77² + 41² - 2*77*41*cos 79
a² = 6405
a = 80 (rounded)

Now find the value of s:

s = 1/2(80 + 77 + 41) = 99

Find the area:

A = $\sqrt{99(99 - 80)(99 - 77)(99 - 41)}$ = $\sqrt{2400156}$ = 1549.24

6 0

3 years ago

Read 2 more answers

If the area of a rectangle is 24a2b and the length is 8ab2, what would be the width of the rectangle, given that width is found

Goshia [24]

(24ba^2)/(8ab^2)=3a/b=width

5 0

4 years ago

Listed below are body temperatures from five different subjects measured at 8 AM and again at 12 AM. Find the values of d overba

neonofarm [45]

Answer:

$(a)\bar {d} =-0.34\\(b)s_{d}=0.073$

Step-by-step explanation:

$\left|\begin{array}{c|c|c}$Temperature at 8 AM$&$Temperature at 12 AM$&$Difference$\\---------&--------&------\\97.8&98.3&-0.5\\98.7&99.3&-0.6\\97.3&97.6&-0.3\\97.5&97.4& 0.1\\97.2&97.6&-0.4\end{array}\right|$

(b)

$\bar {d} $ is the sample mean of all the difference in temperature$\\\bar {d} =\dfrac{-0.5-0.6-0.3+0.1-0.4}{5} =\dfrac{-1.7}{5} =-0.34$

$(-0.5-(-0.34))^{2}=0.0256\\(-0.6-(-0.34))^{2}=0.0676\\(-0.3-(-0.34))^{2}=0.0016\\(0.1-(-0.34))^{2}=0.1936\\(-0.4-(-0.34))^{2}=0.0036\\s_{d} =\dfrac{0.0256+0.0676+0.0016+0.1936+0.0036}{5-1}=\dfrac{0.292}{4}\\s_{d}=0.073$

(d) $\mu_d$ represents the mean of the difference in body temperature at 8AM and 12AM of the population of all people.

6 0

4 years ago