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3 years ago

How would you calculate the area and perimeter of 3yd 6yd 5yd 10yds

Mathematics

1 answer:

spayn [35]3 years ago

3 0

Okay, start with the smaller square, you know the side length is 3 and that the width is 5, multiply this together:
3 * 5 = 15

That's the smaller left side of this shape, now the bigger side, divide the 10 yards at the bottom by 2 to properly represent the other side then multiply this by the 6 over there.
5 * 6 = 30

Conclusion for Area: After adding the results the answer should be 45.

The perimeter is solely adding all the sides together. You have a 10, two 5's, one 6, and two 3's. Add this together:

10 + 5 + 5 + 6 + 3 + 3 = 32

Conclusion for Perimeter: So the perimeter would be 32.

I hope this helps in someway, have a great rest of your day! ^ ^
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Which scenario best matches the inequality x > 3? A. Joanna needs to be in bed by 10 p.M. It's now just past 7 p.M. How many

goldfiish [28.3K]

Answer:

Step-by-step explanation:

If Ricardo started studying at 4 p.m and for sure will not stop until after 7 p.m, that means that he will be studying for 3 hours at least, and maybe even longer. So the time he will spend studying is at least 3 hours.

Therefore, x>3, where x is the time Ricardo will spend studying.

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3 years ago

The top of a table is a parallelogram with a base of 28 cm and a height of 20 cm. Patty wants to cover the table with tiles that

suter [353]

The area of the top of the table is 28 × 20 = 560 cm2. The area of Patty’s tiles is 8 cm2 each and the area of Jake’s tiles is 9 cm2 each. Patty will need 560 ÷ 8, or 70, tiles. Jake will need 560 ÷ 9, or about 63, tiles. So, Patty will use about 7 more tiles.

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3 years ago

Read 2 more answers

The angle of depression from the top of a cruise ship to the top of a sailboat is 22. Sitting above water, the cruise ship is 23

dusya [7]

Answer:

The distance between the cruise ship and the sail boat is 517 feet.

Step-by-step explanation:

The cruise ship is 236 feet tall, and the sailboat is 236 tall; this means the distance between the top of the cruise ship and the top of the sail boat is

236 feet - 27 feet = 209 feet.

We also know that the angel of depression from the top of the cruise ship to the bottom of the cruise ship is 22°. This forms a right triangle as shown in the figure attached.

Now from trigonometry we get:

$tan \: \theta = \dfrac{209}{d}$

$d= \dfrac{209}{tan\:\theta }$

$\boxed{d=517ft}$

The distance between the cruise ship and the sailboat is 517 feet.

3 0

3 years ago

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Gre4nikov [31]

Answer:

93° (to the nearest degree)

Step-by-step explanation:

sum of the interior angles of a triangle = 180°

Find angle C first, then subtract angle C and angle B from 180° to find angle A.

Use the sine rule $\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}$ to find angle C:

Therefore,

$\frac{sinB}{b}=\frac{sinC}{c}$

$\frac{sin38}{9}=\frac{sinC}{11}$

angle C = 48.80523914...°

Angle A = 180 - 38 - 48.80523914...°

= 93.19476086..°

= 93° (to the nearest degree)

7 0

2 years ago

A marketing research firm wishes to compare the prices charged by two supermarket chains—Miller’s and Albert’s. The research fir

AVprozaik [17]

Answer:

a) $F=\frac{s^2_2}{s^2_1}=\frac{1.84^2}{1.4^2}=1.727 \approx 1.73$

b) For this case since we have a two tailed test we have two critical values, and we can find the two values since we need that on each tail the area would be $\alpha/2 =0.025$ , the distribution is the F with 9 df for the numerator and denominator, and we can use the following excel codes to find the critical values:

"=F.INV(0.025,9,9)"

"=F.INV(1-0.025,9,9)"

The two critical values are 0.248 and 4.026, so the rejection zone would be: $F>4.026 \cup F$ since our calculated value is not on the rejection zone we fail to rejec the null hypothesis

Step-by-step explanation:

Data given and notation

$n_1 = 10$ represent the sampe size for the Miller's Stores

$n_2 =10$ represent the sample size for the Albert's stores

$\bar X_1 =121.92$ represent the sample mean for Miller's store

$\bar X_2 =114.81$ represent the sample mean for Albert's store

$s_1 = 1.4$ represent the sample deviation for the Miller's store

$s_2 = 1.84$ represent the sample deviation for the Albert's stores

$s^2_2 = 12.25$ represent the sample variance for the utility stocks

$\alpha=0.05$ represent the significance level provided

Confidence =0.95 or 95%

F test is a statistical test that uses a F Statistic to compare two population variances, with the sample deviations s1 and s2. The F statistic is always positive number since the variance it's always higher than 0. The statistic is given by:

$F=\frac{s^2_2}{s^2_1}$

Solution to the problem

System of hypothesis

We want to test if the variation for th two groups is the same, so the system of hypothesis are:

H0: $\sigma^2_1 = \sigma^2_2$

H1: $\sigma^2_1 \new \sigma^2_2$

a) Calculate the statistic

Now we can calculate the statistic like this:

$F=\frac{s^2_2}{s^2_1}=\frac{1.84^2}{1.4^2}=1.727 \approx 1.73$

Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have $n_1 -1 =10-1=9$ and for the denominator we have $n_2 -1 =10-1=9$ and the F statistic have 9 degrees of freedom for the numerator and 9 for the denominator. And the P value is given by:

P value

$p_v =2*P(F_{9,9}>1.727)=0.428$

And we can use the following excel code to find the p value:"=2*(1-F.DIST(1.727,9,9,TRUE))"

b) Critical value

For this case since we have a two tailed test we have two critical values, and we can find the two values since we need that on each tail the area would be $\alpha/2 =0.025$ , the distribution is the F with 9 df for the numerator and denominator, and we can use the following excel codes to find the critical values:

"=F.INV(0.025,9,9)"

"=F.INV(1-0.025,9,9)"

The two critical values are 0.248 and 4.026, so the rejection zone would be: $F>4.026 \cup F$ since our calculated value is not on the rejection zone we fail to rejec the null hypothesis

Since our calcu

Conclusion

Since the $p_v > \alpha$ we have enough evidence to FAIL to reject the null hypothesis. And we can say that we don't have enough evidence to conclude that the two deviations are different at 5% of significance.

8 0

3 years ago