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Bingel [31]
1 year ago
9

32.15 is written correctly in EXPANDED FORM using FRACTIONS

Mathematics
1 answer:
marshall27 [118]1 year ago
5 0

Answer:

Step-by-step explanation:

: Read and write decimals to thousandths using base-ten numerals, number names, and expanded form, e.g., 347.392 = 3 × 100 + 4 × 10 + 7 × 1 + 3 × (1/10) + 9 × (1/100) + 2 × (1/1000).

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The box which measures 70cm X 36cm X 12cm is to be covered by a canvas. How many meters of canvas of width 80cm would be require
grigory [225]

Answer:

142.2 meters.  

Step-by-step explanation:

We have been given that a box measures 70 cm X 36 cm X 12 cm is to be covered by a canvas.      

Let us find total surface area of box using surface area formula of cuboid.

\text{Total surface area of cuboid}=2(lb+bh+hl), where,

l = Length of cuboid,

b = Breadth of cuboid,

w = Width of cuboid.

\text{Total surface area of box}=2(70\cdot36+36\cdot 12+12\cdot 70)

\text{Total surface area of box}=2(2520+432+840)

\text{Total surface area of box}=2(3792)

\text{Total surface area of box}=7584

Therefore, the total surface area of box will be 7584 square cm.  

To find the length of canvas that will cover 150 boxes, we will divide total surface area of 150 such boxes by width of canvass as total surface area of canvas will also be the same.

\text{Width of canvas* Length of canvass}=\text{Total surface area of 150 boxes}

80\text{ cm}\times\text{ Length of canvass}=150\times 7584\text{cm}^2

\text{ Length of canvass}=\frac{150\times 7584\text{ cm}^2}{80\text{ cm}}

\text{ Length of canvass}=\frac{1137600\text{ cm}^2}{80\text{ cm}}

\text{ Length of canvass}=14220\text{ cm}

Let us convert the length of canvas into meters by dividing 14220 by 100 as 1 meter equals to 100 cm.

\text{ Length of canvass}=\frac{14220\text{ cm}}{100\frac{cm}{m}}

\text{ Length of canvass}=\frac{14220\text{ cm}}{100\frac{cm}{m}}

\text{ Length of canvass}=\frac{14220\text{ cm}}{100}\times\frac{m}{cm}

\text{ Length of canvass}=142.20\text{ m}

Therefore, 142.2 meters of canvas of width 80 cm required to cover 150 such boxes.

5 0
2 years ago
What is the greatest possible percent error in calculating the volume of a box measured to the nearest inch with sides of 5 in.,
Mashutka [201]
The answer is 20 percent
7 0
1 year ago
Read 2 more answers
Suppose the area that can be painted using a single can of spray paint is slightly variable and follows a nearly normal distribu
Svetllana [295]

Answer:

Step-by-step explanation:

Hello!

You have the variable

X: Area that can be painted with a can of spray paint (feet²)

The variable has a normal distribution with mean μ= 25 feet² and standard deviation δ= 3 feet²

since the variable has a normal distribution, you have to convert it to standard normal distribution to be able to use the tabulated accumulated probabilities.

a.

P(X>27)

First step is to standardize the value of X using Z= (X-μ)/ δ ~N(0;1)

P(Z>(27-25)/3)

P(Z>0.67)

Now that you have the corresponding Z value you can look for it in the table, but since tha table has probabilities of P(Z, you have to do the following conervertion:

P(Z>0.67)= 1 - P(Z≤0.67)= 1 - 0.74857= 0.25143

b.

There was a sample of 20 cans taken and you need to calculate the probability of painting on average an area of 540 feet².

The sample mean has the same distribution as the variable it is ariginated from, but it's variability is affected by the sample size, so it has a normal distribution with parameners:

X[bar]~N(μ;δ²/n)

So the Z you have to use to standardize the value of the sample mean is Z=(X[bar]-μ)/(δ/√n)~N(0;1)

To paint 540 feet² using 20 cans you have to paint around 540/20= 27 feet² per can.

c.

P(X≤27) = P(Z≤(27-25)/(3/√20))= P(Z≤2.98)= 0.999

d.

No. If the distribution is skewed and not normal, you cannot use the normal distribution to calculate the probabilities. You could use the central limit theorem to approximate the sampling distribution to normal if the sample size was 30 or grater but this is not the case.

I hope it helps!

4 0
2 years ago
The aquarium at Sea Critters Depot contains 140 fish. Eighty of these fish are green swordtails (44 female and 36 male) and 60 a
maxonik [38]

Given that:

Total number of fish = 140

Fish are green swordtails female = 44

Fish are green swordtails male = 36

Fish are orange swordtails female = 36

Fish are orange swordtails male = 24

Solution:

A. We have to find the probability that the selected fish is a green swordtail.

\text{P(green swordtail)}=\dfrac{\text{Total green swordtail fish}}{\text{Total fish}}

\text{P(green swordtail)}=\dfrac{80}{140}

\text{P(green swordtail)}=\dfrac{4}{7}

Therefore, the probability that the selected fish is a green swordtail is \dfrac{4}{7}.

B.  We have to find the probability that the selected fish is male.

\text{P(Male fish)}=\dfrac{\text{Total male fish}}{\text{Total fish}}

\text{P(Male fish)}=\dfrac{36+24}{140}

\text{P(Male fish)}=\dfrac{60}{140}

\text{P(Male fish)}=\dfrac{3}{7}

Therefore, the probability that the selected fish is a male, is \dfrac{3}{7}.

C. We have to find the probability that the selected fish is a male green swordtail.

\text{P(Male green swordtail)}=\dfrac{\text{Total male green swordtail fish}}{\text{Total fish}}

\text{P(Male green swordtail)}=\dfrac{36}{140}

\text{P(Male green swordtail)}=\dfrac{9}{35}

Therefore, probability that the selected fish is a male green swordtail is \dfrac{9}{35}.

D.

We have to find the probability that the selected fish is either a male or a green swordtail.

\text{P(Male or green swordtail)}=\dfrac{\text{Total male or green swordtail fish}}{\text{Total fish}}

\text{P(Male or green swordtail)}=\dfrac{44+36+24}{140}

\text{P(Male or green swordtail)}=\dfrac{96}{140}

\text{P(Male or green swordtail)}=\dfrac{24}{35}

Therefore, the probability the selected fish is either a male or a green swordtail is \dfrac{24}{35}.

4 0
2 years ago
in a scale drawing of an apartment, 1 centimeter represents 2 3/4 feet. If the length of the kitchen in 4 1/2 cm on the scale dr
kirill [66]

It is 12.375 feet long.

3 0
1 year ago
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