Naimal averages 80km/h driving her car.

Shirl is planting a 4-foot flower border around the edge of her yard. She drew this diagram to show how it should look.

Lostsunrise [7]

Two rectangles, one lying completely within the other. The area between the rectangles is shaded. Two adjacent sides of the larger rectangle are labeled sixty feet and forty feet. The shaded area is labeled four foot border.

Area of larger rectangle = 60 ft * 40 ft = 2400 ft²
Area of smaller rectangle = (60-4)ft * (40-4)ft = 56 ft * 36 ft = 2,016 ft²

Area of the border = 2,400 ft² - 2,016 ft² = 384 ft²

7 0

3 years ago

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Why is the power in the united states dived between several branches

Bingel [31]

Answer:

To ensure that they all have the equal amount of power; to make sure one branch doesn't have more power than the others.

6 0

3 years ago

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4

o-na [289]

Answer:

The measure of exterior angle is 6 is 122

4 0

3 years ago

(sin teta + sec teta)^ + (cos teta+ cosec teta )^ = (1 + sec x cosec)^

ArbitrLikvidat [17]

$(sin\theta+sec\theta)^2+(cos\theta+cosec\theta)^2=(1+sec\theta\ cosec\theta)^2\\\\L=sin^2\theta+2sin\theta\cdot\frac{1}{cos\thewta}+\frac{1}{cos^2\theta}+cos^2\theta+2cos\theta\cdot\frac{1}{sin\theta}+\frac{1}{sin^2\theta}\\\\=(sin^2\theta+cos^2\theta)+\frac{2sin\theta}{cos\theta}+\frac{2cos\theta}{sin\theta}+\frac{1}{cos^2\theta}+\frac{1}{sin^2\theta}$

$=1+\frac{2sin^2\theta+2cos^2\theta}{sin\theta\ cos\theta}+\frac{sin^2\theta+cos^2\theta}{sin^2\theta\ cos^2\theta}\\\\=1+\frac{2(sin^2\theta+cos^2\theta)}{sin\theta\ cos\theta}+\frac{1}{sin^2\theta\ cos^2\theta}\\\\=1+\frac{2}{sin\theta\ cos\theta}+\frac{1}{sin^2\theta\ cos^2\theta}\\\\=1^2+2sec\theta\ cosec\theta+sec^2\theta\ cosec^2\theta\\\\=(1+sec\theta\ cosec\theta)^2=R$

4 0

3 years ago

A random sample of n = 9 structural elements is tested for comprehensive strength. We know the true mean comprehensive strength

irinina [24]

Answer:

Probability that the sample mean comprehensive strength exceeds 4985 psi is 0.99999.

Step-by-step explanation:

We are given that a random sample of n = 9 structural elements is tested for comprehensive strength. We know the true mean comprehensive strength μ = 5500 psi and the standard deviation is σ = 100 psi.

Let $\bar X$ = sample mean comprehensive strength

The z-score probability distribution for sample mean is given by;

Z = $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean comprehensive strength = 5500 psi

$\sigma$ = standard deviation = 100 psi

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the sample mean comprehensive strength exceeds 4985 psi is given by = P( $\bar X$ > 4985 psi)

P( $\bar X$ > 4985 psi) = P( $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{4985-5500}{\frac{100}{\sqrt{9} } }$ ) = P(Z > -15.45) = P(Z < 15.45)

= 0.99999

Since in the z table the highest critical value of x for which a probability area is given is x = 4.40 which is 0.99999, so we assume that our required probability will be equal to 0.99999.

4 0

3 years ago