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3 years ago

Plz help me on 15 plz thanks if you help

Mathematics

1 answer:

Nitella [24]3 years ago

8 0

Units are simplified first when measurements are multiplied because the units can sometime be very large or small and do not have to be exact when measurements are precise.

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The long-distance calls made by the employees of a company are normally distributed with a mean of 6.3 minutes and a standard de

Sergio [31]

To solve this problem, we use the formula:

z = (x – u) / s

where z is the z score value which can be obtained from the tables, x is the sample value, u is the mean = 6.3 min, and s is the std dev = 2.2 min

at P value = 0.90, the z = 1.28, finding for x:

x = z s + u

x = 1.28 * 2.2 + 6.3

x = 9.116

at P value = 1.0, the z = 3.49, finding for x:

x = z s + u

x = 3.49 * 2.2 + 6.3

x = 13.978 ~ 14

Therefore the longest 10% calls last about 9.1 minutes to 14 minutes

3 0

3 years ago

PLS HELP SHOW ALL YOUR WORKING OUT AND THE CORRECT ANSWER WILL RECIEVE BRAINLIEST

EastWind [94]

So the length is 4*28=112mm
The width would be calculated like this:

you draw an equilateral triangle in the middle with sides or 2r(56mm). The based on pythagorean theorem u know that
h=rad(56^2-28^2)=48.497
28+28+h=104.497

The area would be S=l*w which is 11703.7113mm

5 0

3 years ago

Suppose the half-life of a material is 10 days. You have one 1 kg of the material today. How much of the material would you have

boyakko [2]

Answer:

0.50 kg of the material would be left after 10 days.

0.25 kg of the material would be left after 20 days.

Step-by-step explanation:

We have been given that the half-life of a material is 10 days. You have one 1 kg of the material today. We are asked to find the amount of material left after 10 days and 20 days, respectively.

We will use half life formula.

$A=a\cdot(\frac{1}{2})^{\frac{t}{h}}$ , where,

A = Amount left after t units of time,

a = Initial amount,

t = Time,

h = Half-life.

$A=1\text{ kg}\cdot(\frac{1}{2})^{\frac{10}{10}}$

$A=1\text{ kg}\cdot(\frac{1}{2})^{1}$

$A=1\text{ kg}\cdot\frac{1}{2}$

$A=0.5\text{ kg}$

Therefore, amount of the material left after 10 days would be 0.5 kg.

$A=1\text{ kg}\cdot(\frac{1}{2})^{\frac{20}{10}}$

$A=1\text{ kg}\cdot(\frac{1}{2})^{2}$

$A=1\text{ kg}\cdot\frac{1^2}{2^2}$

$A=1\text{ kg}\cdot\frac{1}{4}$

$A=0.25\text{ kg}$

Therefore, amount of the material left after 20 days would be 0.25 kg.

6 0

3 years ago

WHAT IS 5 TIMES 5 IN MATH

sergij07 [2.7K]

Answer:

that is 25

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

A line with a slope of 1/8 passes through the points (u, -9) and (6, -8). what is the value of u?

Pie

Answer:
u = -2

Explanation:
Slope of the line can be calculated using the following rule:
slope = (y2-y1) / (x2-x1)

We are given :
slope = 1/8
point (u,-9) represent (x1,y1)
point (6,-8) represent (x2,y2)

Substitute with the givens in the above equation and solve for u as follows:
slope = (y2-y1) / (x2-x1)
1/8 = (-8--9) / (6-u)
1/8 = (1) / (6-u)
1* (6-u) = 1*8
6-u = 8
u = 6-8
u = -2

Hope this helps :)

6 0

3 years ago