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Evgen [1.6K]
2 years ago
5

Malik's arts and crafts shop has a bolt of crimson velvet 32 meters long. A customer came in during the morning and bought 120 c

entimeters of the velvet. Another customer bought an additional 14 meters during the afternoon. How much crimson velvet does Malik's shop have left? Write the answer in centimeters.
Mathematics
1 answer:
kaheart [24]2 years ago
6 0
Mil seiscientos ochenta,centímetros; le quedan a La tienda Malik’s. 3200-1520=1680 cm
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An 8-pound pack of ground beef sells for $15.04. What is the cost per pound of the ground beef
Alex17521 [72]

Answer:

$1.88

Step-by-step explanation:

$15.04/8 pounds = $1.88

6 0
2 years ago
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Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One common ch
enot [183]

Answer:

y=\frac{-7t^2+22t-7}{7t-22}

Step-by-step explanation:

We are given that

Initial value problem

y'=(t+y)^2-1, y(3)=4

Substitute the value z=t+y

When t=3 and y=4 then

z=3+4=7

y'=z^2-1

Differentiate z w.r.t t

Then, we get

\frac{dz}{dt}=1+y'

z'=1+z^2-1=z^2

z^{-2}dz=dt

Integrate on both sides

-\frac{1}{z}dz=t+C

z=-\frac{1}{t+C}

Substitute t=3 and z=7

Then, we get

7=-\frac{1}{3+C}

21+7C=-1

7C=-1-21=-22

C=-\frac{22}{7}

Substitute the value of C then we get

z=-\frac{1}{t-\frac{22}{7}}

z=\frac{-7}{7t-22}

y=z-t

y=\frac{-7}{7t-22}-t

y=\frac{-7-7t^2+22t}{7t-22}

y=\frac{-7t^2+22t-7}{7t-22}

8 0
2 years ago
The area of a square concrete slab is 20 square feet. Find the side length of the square concrete slab if the area is increase b
guapka [62]

The side length of the square concrete slab if the area is increased by 25% is 5feet

The formula for calculating the area of a square is expressed as:

A =  L² where:

L is the side length of the square

Given the area of the square concrete slab = 20 square feet

20 = L²

L =√20

If the area is increased  by 25%, new area will be:

An = 20 + (0.25*20)

An = 20 + 5

An = 25 sq.ft

Get the new length

An = Ln²

25 =   Ln²

Ln = √25

Ln = 5feet

Hence the side length of the square concrete slab if the area is increased by 25% is 5feet

Learn more here: brainly.com/question/11300671

7 0
2 years ago
A study was performed on green sea turtles inhabiting a certain area.​ Time-depth recorders were deployed on 6 of the 76 capture
polet [3.4K]

Answer:

One can be 99% confident the true mean shell length lies within the above interval.

The population has a relative frequency distribution that is approximately normal.

Step-by-step explanation:

We are given that Time-depth recorders were deployed on 6 of the 76 captured turtles. These 6 turtles had a mean shell length of 51.3 cm and a standard deviation of 6.6 cm.  

The pivotal quantity for a 99% confidence interval for the true mean shell length is given by;

                    P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean shell length = 51.3 cm

             s = sample standard deviation = 6.6 cm

             n = sample of turtles = 6

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Here, \alpha = 1% so  (\frac{\alpha}{2}) = 0.5%. So, the critical value of t at 0.5% significance level and 5 (6-1) degree of freedom is 4.032.

<u>So, 99% confidence interval for</u> \mu  =  51.3 \pm 4.032 \times \frac{6.6}{\sqrt{6} }

                                                         = [51.3 - 10.864 , 51.3 + 10.864]

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The interpretation of the above result is that we are 99% confident that the true mean shell length lie within the above interval of [40.44 cm, 62.16 cm].

The assumption about the distribution of shell lengths must be true in order for the confidence interval, part a, to be valid is that;

C. The population has a relative frequency distribution that is approximately normal.

This assumption is reasonably satisfied as the data comes from the whole 76 turtles and also we don't know about population standard deviation.

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You have everything right. If you look, it says Y is distance from home, and X is the time they spent driving.
I took the test and made a 100%
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2 years ago
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