Rou<br>Round 66.681 to the nearest ten thousand.

A gardening center is planting trees for sale. Until the trees are fully grown, employees need to put rope

Zepler [3.9K]

Answer: 46ft

Step-by-step explanation:

You take the area and find it's square root then multiply it by how many sides there are to the figure in this case it is 4 sides so you multiply it by 4 getting 46

6 0

3 years ago

Ill mark btainlist pllssss helpp

Hatshy [7]

Answer:

1cm-----80000m

1cm-----8000000cm

1cm×3.275-----80km×3.275

3.275cm

3 0

3 years ago

If a store had 330 video games, and 47% of the games were for Xbox, and 53% were for ps4, then how many xbox and ps4 games are i

Bingel [31]

Answer:

The number of games for Xbox is about 155

The number of games for PS4 is about 175

Step-by-step explanation:

Let us find the numbers of games for Xbox and for PS4 game

∵ There are 330 video games in the store

∵ 47% of them were for Xbox

- Multiply 47% by 330

∴ The number of games for Xbox = 47% × 330

∵ 47% = $\frac{47}{100}$

∴ The number of games for Xbox = $\frac{47}{100}$ × 330

∴ The number of games for Xbox = 155. 1

∴ The number of games for Xbox ≅ 155

∵ 53% of them were for PS4

- Multiply 53% by 330

∴ The number of games for PS4 = 53% × 330

∵ 53% = $\frac{53}{100}$

∴ The number of games for PS4 = $\frac{53}{100}$ × 330

∴ The number of games for PS4 = 174.9

∴ The number of games for PS4 ≅ 175

4 0

3 years ago

Nine out of ten students prefer math class over lunch. How many students do not prefer math if 200 students were asked?

fredd [130]

So, 9 out of 10 students prefer class, so, obviously, it would not be 100. It would be 150. So in conclusion, 50 students prefer lunch over math class, whilst 150 other students prefer math class over lunch.

3 0

3 years ago

he owner of a local supermarket wants to estimate the difference between the average number of gallons of milk sold per day on w

stellarik [79]

Answer:

90% confidence interval is ( -149.114, -62.666 )

Step-by-step explanation:

Given the data in the question;

Sample 1 Sample 2

x"₁ = 259.23 x"₂ = 365.12

s₁ = 34.713 s₂ = 48.297

n₁ = 5 n₂ = 10

With 90% confidence interval for μ₁ - μ₂ { using equal variance assumption }

significance level ∝ = 1 - 90% = 1 - 0.90 = 0.1

Since we are to assume that variance are equal and they are know, we will use pooled variance;

Degree of freedom DF = n₁ + n₂ - 2 = 5 + 10 - 2 = 13

Now, pooled estimate of variance will be;

$S_p^2$ = [ ( n₁ - 1 )s₁² + ( n₂ - 1)s₂² ] / [ ( n₁ - 1 ) + ( n₂ - 1 ) ]

we substitute

$S_p^2$ = [ ( 5 - 1 )(34.713)² + ( 10 - 1)(48.297)² ] / [ ( 5 - 1 ) + ( 10 - 1 ) ]

$S_p^2$ = [ ( 4 × 1204.9923) + ( 9 × 2332.6 ) ] / [ 4 + 9 ]

$S_p^2$ = [ 4819.9692 + 20993.4 ] / [ 13 ]

$S_p^2$ = 25813.3692 / 13

$S_p^2$ = 1985.64378

Now the Standard Error will be;

$S_{x1-x2$ = √[ ( $S_p^2$ / n₁ ) + ( $S_p^2$ / n₂ ) ]

we substitute

$S_{x1-x2$ = √[ ( 1985.64378 / 5 ) + ( 1985.64378 / 10 ) ]

$S_{x1-x2$ = √[ 397.128756 + 198.564378 ]

$S_{x1-x2$ = √595.693134

$S_{x1-x2$ = 24.4068

Critical Value = $t_{\frac{\alpha }{2}, df$ = $t_{0.05, df=13$ = 1.771 { t-table }

So,

Margin of Error E = $t_{\frac{\alpha }{2}, df$ × [ ( $S_p^2$ / n₁ ) + ( $S_p^2$ / n₂ ) ]

we substitute

Margin of Error E = 1.771 × 24.4068

Margin of Error E = 43.224

Point Estimate = x₁ - x₂ = 259.23 - 365.12 = -105.89

So, Limits of 90% CI will be; x₁ - x₂ ± E

Lower Limit = x₁ - x₂ - E = -105.89 - 43.224 = -149.114

Upper Limit = x₁ - x₂ - E = -105.89 + 43.224 = -62.666

Therefore, 90% confidence interval is ( -149.114, -62.666 )

3 0

3 years ago

RouRound 66.681 to the nearest ten thousand.​

RouRound 66.681 to the nearest ten thousand.