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3 years ago

A right rectangular prism is 5 1/4 in. wide, 12 1 2 in. long, and 4 in. tall.

Mathematics

1 answer:

lozanna [386]3 years ago

7 0

Answer: $V = 262\frac{1}{2}$ $cm^{3}$

Step-by-step explanation:

$V = LWH$

$V = 5\frac{1}{4}$ X $12\frac{1}{2}$ X 4

$V = \frac{21}{4}$ X $\frac{25}{2}$ X $\frac{4}{1}$

$V = \frac{525}{2}$

$V = 262\frac{1}{2}$ $cm^{3}$

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Rectangle A is a scale drawing of Rectangle B and has 25% of its area. If rectangle A has side lengths of 4 cm and 5 cm , what a

SpyIntel [72]

Answer:

8 cm and 10 cm

Step-by-step explanation:

Hello, <em> </em>I can help you with this.

Step 1

According to the question there are two rectangles A and B,

Rectangle A is a scale drawing of Rectangle B and has 25% of its area

in other words

$Area_{A}=0.25*Area_{B} (Equation\ 1)\\$

Step 2

Let

Rectangle A

length (1)= 4 cm

length (2)= 5 cm

$Area_{A}=4\ cm * 5\ cm\\Area_{A}=20\ cm^{2}$

put this value into equation 1

$Area_{A}=0.25*Area_{B} (Equation\ 1)\\\\20\ cm^{2} =0.25*Area_{B} \\divide\ each\ side\ by\ 0.25\\\frac{20\ cm^{2} }{0.25}=\frac{0.25}{0.25}*Area_{B}\\ Area_{B}=80\ cm^{2}$

Now, we know the area of rectangle B, to know its length we need to formule other equation

Step 3

$Area_{B}=80\ cm^{2}\\length (1B)*length (2B)=80\ cm^{2} (equation\ 2)\\$

the ratio between the lengths must be constant, so the ratio of A must be equal to ratio in B, then

$\frac{length(1A)}{length(2A)}=\frac{length(1B)}{length(2B)} \\\\\\frac{4}{5}= \frac{length(1B)}{length(2B)}\\0.8=\frac{length(1B)}{length(2B)}\\length(1B)=0.8*length(2B) (Equation 3)$

Step three

using Eq 1 and Eq 2 find the lengths

put the value of length(1B) into equation (2)

$length (1B)*length (2B)=80\ cm^{2} (equation\ 2)\\\(0.8*length(2B)) (*length (2B)=80\ cm^{2} \\\\0.8*(length (2B))^{2} =80\ cm^{2}\\(length (2B))^{2} =\frac{80\ cm^{2}}{0.8} \\(length (2B))^{2}=100\\\sqrt{(length (2B))^{2}}=\sqrt{100\ cm^{2}} \\ length (2B)=10\ cm$

Now, put the value of length(2B) into equation 3 to know length (1B)

$length(1B)=0.8*length(2B)\\length(1B)=0.8*10\ cm\\length(1B)=8 cm$

I really hope this helps you, have a great day.

6 0

3 years ago

Can someone solve this please

Elanso [62]

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7 0

3 years ago

Sophia exchanged 1000 U.S. dollars for the South African currency which is called the rand. The exchange rate was 8.05 rand to 1

saw5 [17]

Answer:

Sophia gets 7,150

Step-by-step explanation:

6 0

3 years ago

124 -115 is close or far apart and ,what is the strategy to get the answer

serious [3.7K]

Its not far apart if it 15 or more yes it far

4 0

3 years ago

It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per nig

Kitty [74]

Answer: No , at 0.05 level of significance , we have sufficient evidence to reject the claim that LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average.

Step-by-step explanation:

Let $\mu$ denotes the average hours of sleep per night.

As per given , we have

$H_0:\mu=7\\H_a:\mu$

, since $H_a$ is left-tailed and population standard deviation is unknown, so the test is a left-tailed t -test.

Also , it is given that ,

Sample size : n= 22

Sample mean : $\overline{x}=7.24$

Sample standard deviation : s= 1.93

Test statistic : $t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}$

i.e. $t=\dfrac{7.24-7}{\dfrac{1.93}{\sqrt{22}}}\approx0.58$

For significance level $\alpha=0.05$ and degree of freedom 21 (df=n-1),

Critical t-value for left-tailed test= $t_{\alpha, df}=t_{0.05,21}=- 1.7207$

Decision : Since the test statistic value (0.58) > critical value 1.7207, it means we are failed to reject the null hypothesis .

[Note : When $|t_{cal}|>|t_{cri}|$ , then we accept the null hypothesis.]

Conclusion: We have sufficient evidence to reject the claim that LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average.

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3 years ago