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3 years ago

5

I have a 15ft x 36in (4.57m x 91 cm) pool and the mats i wanna buy are 20.5-in x 20.5-in. How many mats would i need to put unde

r the pool?

1 answer:

yuradex [85]3 years ago

4 0

Answer:

you would need 4 matts

Step-by-step explanation:

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Help based offf order of operations please help

JulijaS [17]

Answer:

The answer for :

$h. \: \: \: \frac{5}{6}$

$i. \: \: \: \frac{35}{32}$

$k. \: \: \: \frac{-29}{20}$

$l. \: \: \: \frac{13}{15}$

Step-by-step explanation:

Question h:

$\frac{2}{3} + ( \frac{1}{3} \times \frac{1}{2} )$

$= \frac{2}{3} + \frac{1}{6}$

$= \frac{2 \times 2}{3 \times 2} + \frac{1}{6}$

$= \frac{4}{6} + \frac{1}{6}$

$= \frac{5}{6}$

Question i:

$\frac{7}{8} + \frac{1}{4} \times ( \frac{3}{2} - \frac{5}{8} )$

$= \frac{7}{8} + \frac{1}{4} \times ( \frac{3 \times 4}{2 \times 4} - \frac{5}{8} )$

$= \frac{7}{8} + \frac{1}{4} \times ( \frac{12}{8} - \frac{5}{8} )$

$= \frac{7}{8} + (\frac{1}{4} \times \frac{7}{8} )$

$= \frac{7}{8} + \frac{7}{32}$

$= \frac{7 \times 4}{8 \times 4} + \frac{7}{32}$

$= \frac{28}{32} + \frac{7}{32}$

$= \frac{35}{32}$

Question k:

$\frac{3}{4} - ( \frac{12}{7} \div \frac{12}{21} ) + \frac{4}{5}$

$= \frac{3}{4} - ( \frac{12}{7} \times \frac{21}{12} ) + \frac{4}{5}$

$= \frac{3}{4} - \frac{3}{1} + \frac{4}{5}$

$= \frac{3 \times 5}{4 \times 5} - \frac{3 \times 20}{1 \times 20} + \frac{4 \times 4}{5 \times 4}$

$= \frac{15}{20} - \frac{60}{20} + \frac{16}{20}$

$= - \frac{29}{20}$

Question l:

$\frac{5}{2} \times ( \frac{2}{3} - \frac{1}{5} ) - ( \frac{2}{5} \div \frac{4}{3} )$

$= \frac{5}{2} \times ( \frac{2 \times 5}{3 \times 5} - \frac{1 \times 3}{5 \times 3} ) - ( \frac{2}{5} \times \frac{3}{4} )$

$= \frac{5}{2} \times ( \frac{10}{15} - \frac{3}{15} ) - \frac{3}{10}$

$= (\frac{5}{2} \times \frac{7}{15}) - \frac{3}{10}$

$= \frac{7}{6} - \frac{3}{10}$

$= \frac{7 \times 5}{6 \times 5} - \frac{3 \times 3}{10 \times 3}$

$= \frac{35}{30} - \frac{9}{30}$

$= \frac{26}{30}$

$= \frac{13}{15}$

8 0

3 years ago

The arc shown is 140 degrees. Find the measure of angle 1.

Elodia [21]

The angle subtended by two chords is half the enclosed arc measure. If we imagine one of the chords getting shorter and shorter this remains true. In the limit the shrinking chord becomes a tangent, and the angle is still half the arc measure.

Answer: 70 degrees

3 0

3 years ago

Sorry, you guys I put the subject in Math its english! But have a nice day! and please help me!

Vilka [71]

Answer:

disappoint

recall

disgusted

reunited

prepare

5 0

2 years ago

Read 2 more answers

In an analysis of healthcare data, ages have been rounded to the nearest multiple of 5 years. The difference between the true ag

Mashcka [7]

Answer:

The correct answer is D. 0.77

Step-by-step explanation:

$X_i \text{ is the difference between true and reported age. where i = 1,2,3......48 }\\\text{The rounded age are uniformly distributed. So, }X_i = 0\\\\and,\thinspace O_{X_i}^2=\frac{\text{(Upper Bound - Lower Bound})^2}{5\times 2.5\text{ ( Rounded value )}}\\\\O_{X_i}^2=\frac{(2.5 + 2.5)^2}{12}\approx 2.083\\\\O_X_i=1.443$

$\bar{X_{48}}=\frac{X_1+X_2+.......+X_{48}}{48}\\\\\bar{X_{48}}=\frac{\frac{48}{X_i}}{48}=0\\\\O_{\bar{X}_{48}}^2=\frac{48\cdot O_{X_i}^2}{48^2}=\frac{O_{X_i}^2}{48}\\\\\text{So, }\bar{X_i}\text{is approximately normal with mean 0 and standard deviation = }\\\\\frac{1.443}{\sqrt{48}}\approx 0.2083$

$=Pr[\frac{-1}{4}\leq \bar{X}_{48}\leq \frac{1}{4}]\\\\=Pr[\frac{-0.25}{0.2083}\leq Z\leq \frac{0.25}{0.2083}]\\\\=Pr[-1.2\leq Z\leq 1.2]\\=Pr(Z\leq 1.2)-Pr(Z\leq -1.2)\\=Pr(Z\leq 1.2)-(1-Pr(Z\leq 1.2))\\=2\times 0.8849 -1\text{ ( Z value for 1.2 is 0.8849 )}\\\approx 0.77$

Hence, the correct answer is 0.77

7 0

3 years ago

Find the volume and surface area of the composite figure. Give your answer in terms of π. HELP ASAP!!

NARA [144]

Answer:

Part 1) The volume of the composite figure is $620.7\pi\cm^{3}$

Part 2) The surface area of the composite figure is $273\pi\ cm^{2}$

$V=620.7\pi\cm^{3}, S=273\pi\ cm^{2}$

Step-by-step explanation:

Part 1) Find the volume of the composite figure

we know that

The volume of the figure is equal to the volume of a cone plus the volume of a hemisphere

<em>Find the volume of the cone</em>

The volume of the cone is equal to

$V=\frac{1}{3} \pi r^{2} h$

we have

$r=7\ cm$

Applying Pythagoras Theorem find the value of h

$h^{2}=25^{2} -7^{2} \\ \\h^{2}= 576\\ \\h=24\ cm$

substitute

$V=\frac{1}{3} \pi (7)^{2} (24)$

$V=392 \pi\cm^{3}$

<em>Find the volume of the hemisphere</em>

The volume of the hemisphere is equal to

$V=\frac{4}{6}\pi r^{3}$

we have

$r=7\ cm$

substitute

$V=\frac{4}{6}\pi (7)^{3}$

$V=228.7\pi\cm^{3}$

therefore

The volume of the composite figure is equal to

$392 \pi\cm^{3}+228.7\pi\cm^{3}=620.7\pi\cm^{3}$

Part 2) Find the surface area of the composite figure

we know that

The surface area of the composite figure is equal to the lateral area of the cone plus the surface area of the hemisphere

<em>Find the lateral area of the cone</em>

The lateral area of the cone is equal to

$LA=\pi rl$

we have

$r=7\ cm$

$l=25\ cm$

substitute

$LA=\pi(7)(25)$

$LA=175\pi\ cm^{2}$

<em>Find the surface area of the hemisphere</em>

The surface area of the hemisphere is equal to

$SA=2\pi r^{2}$

we have

$r=7\ cm$

substitute

$SA=2\pi (7)^{2}$

$SA=98\pi\ cm^{2}$

Find the surface area of the composite figure

$175\pi\ cm^{2}+98\pi\ cm^{2}=273\pi\ cm^{2}$

4 0

3 years ago