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3 years ago

Write the mixed number -2 and 7/18 as a decimal

Mathematics

1 answer:

Anna71 [15]3 years ago

6 0

Answer:

5.142857

Step-by-step explanation:

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salantis [7]

Answer:

Step-by-step explanation:

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3 years ago

The volumes of two similar solids are 210 m3 and 1,680 m3. The surface area of the larger solid is 856 m2. What is the surface a

Ksju [112]

we know that

$scale \ factor^{3}=\frac {volume\ larger\ solid }{volume\ smaller\ solid}$

Find the value of the scale factor

$volume\ larger\ solid= 1,680\ m^{3} \\ volume\ smaller\ solid= 210\ m^{3}$

substitute the values in the formula

$scale \ factor^{3}=\frac {1,680 }{210}$

$scale \ factor^{3}=8$

$scale \ factor=\sqrt[3]{8} \\ scale \ factor= 2$

Find the surface area of the smaller solid

we know that

$scale \ factor^{2}=\frac {surface\ area\ larger\ solid }{surface\ area\ smaller\ solid}$

$surface\ area\ larger\ solid =856\ m^{2} \\ scale\ factor =2$

$surface\ area\ smaller\ solid= \frac{surface\ area\ larger\ solid}{scale \ factor^{2}}$

substitute the values

$surface\ area\ smaller\ solid= \frac{856}{2^{2}}$

$surface\ area\ smaller\ solid=214\ m^{2} }$

therefore

the answer is

The surface area of the smaller solid is equal to $214\ m^{2}$

3 0

3 years ago

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Sergeeva-Olga [200]

Answer:

2.9×

${10}^{8}$

5 0

2 years ago

Neelam is pouring sand into a clear box to make sand art. The box is a right rectangular prism, and the base of the box is a rig

Komok [63]

This is math right it’s Easter with a tutor

3 0

3 years ago

The Apple, Inc. sales manager for the Chicago West Suburban region is disturbed about the large number of complaints her office

ankoles [38]

Answer:

0.6856

Step-by-step explanation:

$\text{The missing part of the question states that we should Note: that N(108,20) model to } \\ \\ \text{ } \text{approximate the distribution of weekly complaints).]}$

Now; assuming X = no of complaints received in a week

Required:

To find P(77 < X < 120)

Using a Gaussian Normal Distribution ( $\mu =$ 108, $\sigma$ = 20)

Using Z scores:

$Z = \dfrac{77-108}{20} \\\\ Z = -\dfrac{35}{20} \\ \\ Z -1.75$

As a result X = 77 for N(108,20) is approximately equal to to Z = -1.75 for N(0,1)

SO;

$Z = \dfrac{120-108}{20} \\ \\ Z = \dfrac{12}{20}\\ \\ Z = 0.6$

Here; X = 77 for a N(108,20) is same to Z = 0.6 for N(0,1)

Now, to determine:

P(-1.75 < Z < 0.6) = P(Z < 0.6) - P( Z < - 1.75)

From the standard normal Z-table:

P(-1.75 < Z < 0.6) = 0.7257 - 0.0401

P(-1.75 < Z < 0.6) = 0.6856

3 0

3 years ago