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tatiyna
3 years ago
8

HELP ME PLEASE!! MATHEMATICS

Mathematics
1 answer:
ankoles [38]3 years ago
3 0
You show the text of the problem, but you need to show the figure too.
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Greyhounds can reach speeds of 20.1 meters per second. Which conversion factor can be used to find this speed in meters per minu
kicyunya [14]

Answer:

Step-by-step explanation:

20.1metres / seconds

We want to convert to metres /minutes

It is known that 60seconds=1minutes

So therefore, either we multiply 20.1metres/seconds by

60seconds/1minutes

Or

1minutes/60seconds

Multiplying with the above does not change the magnitude of the quantity because it is like we are multiplying by 1.

Since we want to cancel seconds and it is in the denominator, so to do this we need to multiply with the fraction that has the seconds as numerator.

So, we are going to multiply with 60seconds/1minutes

20.1metres/seconds ×60seconds/minutes

1206metres/minutes.

So the correct fraction is the StartFraction 60 seconds Over 1 minute, which is the third option

7 0
3 years ago
Read 2 more answers
The length of a rectangle is 5 more than the width. The perimeter is 120. What is the length, width, and area? Please HELP
Ad libitum [116K]

Answer:

l = 32.5 units, w= 27.5 units, A = 893.75 units²

Step-by-step explanation:

width is w

length is l = 5+w

P = 2( l+w) , substitute l for 5+w

P = 2(5+w+w)

P = 2(5+2w)

P = 10 +4w

P = 120

10 +4w = 120

4w = 120-10

4w = 110

w= 110/4

w= 27.5 units

l = 5+w = 5+ 27.5 = 32.5 units

A = l*w = 27.5 * 32.5 = 893.75 units ²

5 0
2 years ago
??????????????????????????
Umnica [9.8K]

Solution:

<em>Simple Interest = Principal Amount × Rate of Interest/100 × Time</em>

Here, Principal Amount = $6000

Rate of Interest =  6%

Time = 4 years

Simple Interest = 6000 × 6/100 × 4 = <em>$1440</em>

3 0
3 years ago
How many students must be randomly selected to estimate the mean weekly earnings of students at one college? We want 95% confide
kari74 [83]

Answer:

The sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:

CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}

The margin of error of a (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:

MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}

The information provided is:

<em>σ</em> = $60

<em>MOE</em> = $2

The critical value of <em>z</em> for 95% confidence level is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Compute the sample size as follows:

MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}

       n=[\frac{z_{\alpha/2}\times \sigma }{MOE}]^{2}

          =[\frac{1.96\times 60}{2}]^{2}

          =3457.44\\\approx 3458

Thus, the sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.

8 0
3 years ago
(3x^3 + 2x^2 - 2) + (x^2 + 5x - 6)
Alecsey [184]

Answer:

3x^3+3x^2+5x-8

Step-by-step explanation:

8 0
3 years ago
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