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2 years ago

Which represents a side length of a square that has an area of 450 square inches? 15 sqrt 2 in. 15 sqrt 3in. 112.5 in. 115.5 in.

Mathematics

2 answers:

luda_lava [24]2 years ago

7 0

$\sqrt{450}=\sqrt{3^2\cdot 5^2\cdot 2}=15\sqrt{2}$

first choice is correct

Virty [35]2 years ago

3 0

15√2

area = s² = 450 ← s is length of side

to find s take the square root of both sides

s = $\sqrt{450}$ = $\sqrt{2(9)(25)}$

= √2 × √9 × √25 = √2 × 3 × 5 = 15√2

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Which system of equations can you use to find the roots of the equation 2x3 + 4x2 – x + 5 = –3x2 + 4x + 9? y = 2x3 + x2 + 3x +5

Alona [7]

Answer:The answer is y = 2x3 + 4x2 – x + 5 and y = –3x2 + 4x + 9

Roots are both: x=-4, x= -1/2 , x= 1

Proof:

Solve for x over the real numbers:

2 x^3 + 4 x^2 - x + 5 = -3 x^2 + 4 x + 9

Subtract -3 x^2 + 4 x + 9 from both sides:

2 x^3 + 7 x^2 - 5 x - 4 = 0

The left hand side factors into a product with three terms:

(x - 1) (x + 4) (2 x + 1) = 0

Split into three equations:

x - 1 = 0 or x + 4 = 0 or 2 x + 1 = 0

Add 1 to both sides:

x = 1 or x + 4 = 0 or 2 x + 1 = 0

Subtract 4 from both sides:

x = 1 or x = -4 or 2 x + 1 = 0

Subtract 1 from both sides:

x = 1 or x = -4 or 2 x = -1

Divide both sides by 2:

Answer: x = 1 or x = -4 or x = -1/2

Step-by-step explanation:

4 0

1 year ago

Read 2 more answers

Of the 500 sample households in the previous exercise, 7 had three or more large-screen TVs. (a) The percentage of households in

likoan [24]

Answer:

a) The percentage of households in the town with three or more largescreen TVs is estimated as :

The best estimation for the population proportion is :

$\hat p=\frac{7}{500}=0.014$

And that represent the 1.4%.

b) And the 95% confidence interval would be given (0.00370;0.0243).

And the % would be between 0.37% and 2.43%.

Step-by-step explanation:

Data given and notation

n=500 represent the random sample taken

X=7 represent the households with three or more large-screen TVs

$\hat p=\frac{7}{500}=0.014$ estimated proportion of households with three or more large-screen TVs

$\alpha=0.05$ represent the significance level (no given, but is assumed)

z would represent the statistic (variable of interest)

p= population proportion of households with three or more large-screen TVs

Part a

The percentage of households in the town with three or more largescreen TVs is estimated as :

The best estimation for the population proportion is :

$\hat p=\frac{7}{500}=0.014$

And that represent the 1.4%.

Part b

Yes is possible. We hav that $np>10$ and $n(1-p)>10$ so we have the assumption of normality to find the interval.

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.014 - 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.00370$

$0.014 + 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.0243$

And the 95% confidence interval would be given (0.00370;0.0243).

And the % would be between 0.37% and 2.43%.

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2 years ago

At a particular restaurant, each onion ring has 70 calories and each slider has 200 calories. A combination meal with onion ring

crimeas [40]

Answer:

3x + 6y

x = sliders

y = onion rings

Step-by-step explanation:

2:1 ratio of onion rings to sliders, add them 200+70+70 = 340,

then 1020/340 = 3, then input into an equation remembering the ratio of onion rings to sliders

7 0

2 years ago

Find the first, fourth, and eighth terms of the sequence. A(n) = –3 • 2^n–1

Sphinxa [80]

A(n) = –3 • 2⁽ⁿ⁻¹⁾

for n = 1 , A₁ = -3.(2)⁰ = -3
for n = 2 , A₂ = -3.(2)¹ = -6
for n = 3 , A₃ = -3.(2)² = -12
for n = 4 , A₄ = -3.(2)³ = -24
...........................................
for n = 8 , A₈ = -3.(2)⁷ = -384

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2 years ago

Merrill Lynch Securities and Health Care Retirement Inc. are two large employers in downtown Toledo, Ohio. They are considering

ratelena [41]

Answer:

Step-by-step explanation:

We would determine the mean an standard deviation first

Mean = (107 + 92 + 97 + 95 + 105 + 101 + 91 + 99 + 95 + 104)/10 = 98.6

Standard deviation = √(summation(x - mean)/n

n = 10

Summation(x - mean) = (107 - 98.6)^2 + (92 - 98.6)^2 + (97 - 96.6)^2 + (95 - 98.6)^2 + (105 - 98.6)^2 + (101 - 98.6)^2 + (91 - 98.6)^2 + (99 - 98.6)^2 + (95 - 98.6)^2 + (104 - 98.6)^2 = 274

Standard deviation = √(274/10) = 5.23

The confidence interval is used to determine the range of values that could possibly contain a population parameter (population mean)

Confidence interval is written in the form,

(Point estimate - margin of error, Point estimate + margin of error)

The sample mean, x is the point estimate for the population mean.

We will use the t distribution because the sample size is small and the population standard deviation is not given.

Degree of freedom, df = 10 - 1 = 9

α = 1 - 0.99 = 0.01

z α/2 = 0.01/2 = 0.005

The area to the right of z 0.005 is 0.005 and the area to the left of z0.005 is 1 - 0.005 = 0.995. Using the t distribution table,

z = 3.25

Margin of error = z × s/√n

Where

s = sample standard Deviation = 5.23

Margin of error = 3.25 × 5.23/√10 = 5.38

Confidence interval = sample mean(point estimate) ± z × σ/√n

The lower boundary of the confidence interval is

98.6 - 5.38 = 93.22

The upper boundary of the confidence interval is

98.6 + 5.38 = 103.98

We estimate with 99% confidence that the mean weekly child care cost of their employees is between $93.22 and $103.98

Also,

99% of the confidence intervals constructed in this way would contain the true value for the population mean of mean weekly child care cost of their employees.

6 0

2 years ago