I need help urgent<br><br> log10 10² =<br><br> -2<br> 1<br> 2

Simplify 6/4.<br> Thank tou

Angelina_Jolie [31]

Answer:

2 1/2

Step-by-step explanation:

6/4=3/2=1 1/2

5 0

3 years ago

Read 2 more answers

Mr. Cruise drove 70 miles in March he drove seven times as many miles in March as he did in January he drove four times as many

SVEN [57.7K]

70/7= 10 miles in January
10/4 = 2.5 miles in February

5 0

3 years ago

Use differentials to estimate the amount of paint needed (in m3) to apply a coat of paint 0.04 cm thick to a hemispherical dome

pogonyaev

The amount of paint is the volume of paint needed.

The amount of paint needed is 0.065312 cubic meters

<h3>How to determine the amount of paint needed</h3>

The volume of a hemisphere is:

$V = \frac 23\pi r^3$

Differentiate the above equation

$V' = 2\pi r^2 r'$

The above equation represents the estimate of the amount of paint needed.

Where:

r represents the radius (r = 52/2 m)
r' represents the thickness (r' = 0.04 cm)

So, we have:

$V' = 2\pi r^2 r'$

$V' = 2 * 3.14 * (52/2\ m)^2 * (0.04cm)$

$V' = 2 * 3.14 * (26\ m)^2 * (0.04cm)$

Express cm as m

$V' = 2 * 3.14 * (26\ m)^2 * (0.0004m)$

$V' = 0.065312 m^3$

Hence, the amount of paint needed is 0.065312 cubic meters

Read more about volumes at:

brainly.com/question/10171109

3 0

2 years ago

Which measurement is most likely to be used to determine the diameter of the well cap?

ycow [4]

Answer:

B. the second one

Step-by-step explanation:

7 0

3 years ago

Health insurers are beginning to offer telemedicine services online that replace the common office visit. A company provides a v

Veronika [31]

Answer:

$\text {CI} = (60.54, \: 81.46)\\\\$

Therefore, we are 95% confident that actual mean savings for a televisit to the doctor is within the interval of ($60.54 to $81.46)

Step-by-step explanation:

Let us find out the mean savings for a televisit to the doctor from the given data.

Using Excel,

=AVERAGE(number1, number2,....)

The mean is found to be

$\bar{x} = \$71$

Let us find out the standard deviation of savings for a televisit to the doctor from the given data.

Using Excel,

=STDEV(number1, number2,....)

The standard deviation is found to be

$s = \$ 22.35$

The confidence interval is given by

$\text {confidence interval} = \bar{x} \pm MoE\\\\$

Where the margin of error is given by

$$ MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $ \\\\$

Where n is the sample of 20 online doctor visits, s is the sample standard deviation and $t_{\alpha/2}$ is the t-score corresponding to a 95% confidence level.

The t-score is given by is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 20 - 1 = 19

From the t-table at α = 0.025 and DoF = 19

t-score = 2.093

So, the margin of error is

$MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) \\\\MoE = 2.093\cdot \frac{22.35}{\sqrt{20} } \\\\MoE = 2.093\cdot 4.997\\\\MoE = 10.46\\\\$

So the required 95% confidence interval is

$\text {CI} = \bar{x} \pm MoE\\\\\text {CI} = 71 \pm 10.46\\\\\text {CI} = 71 - 10.46, \: 71 + 10.46\\\\\text {CI} = (60.54, \: 81.46)\\\\$

Therefore, we are 95% confident that actual mean savings for a televisit to the doctor is within the interval of ($60.54 to $81.46)

7 0

3 years ago

I need help urgent log10 10² = -2 1 2