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3 years ago

10

A rhombus has a perimeter of 484 feet. What is the length of each side of the rhombus?

1 answer:

LenKa [72]3 years ago

5 0

Each side of the rhombus is 121 feet

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Write -1/2 as a decimal number

Ann [662]

<h2>Answer:</h2>

<u>1/2 is equal to 0.5, so </u><u>-1/2 is -0.5</u><u>.</u>

If you divide -1 by 2, you get -0.5.

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3 years ago

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The amount of money Jerome saves after w weeks is represented by the function f(w)=15w+230

Aliun [14]

15 represents the slope of the function. In other words it means that for every week Jerome earns 15 dollars more. Hope this helps :)

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2 years ago

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Cable Strength: A group of engineers developed a new design for a steel cable. They need to estimate the amount of weight the ca

KatRina [158]

Answer:

95% confidence interval for the mean breaking strength of the new steel cable is [763.65 lb , 772.75 lb].

Step-by-step explanation:

We are given that the engineers take a random sample of 45 cables and apply weights to each of them until they break. The mean breaking weight for the 45 cables is 768.2 lb. The standard deviation of the breaking weight for the sample is 15.1 lb.

Since, in the question it is not specified that how much confidence interval has be constructed; so we assume to be constructing of 95% confidence interval.

Firstly, the Pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean breaking weight = 768.2 lb

s = sample standard deviation = 15.1 lb

n = sample of cables = 45

$\mu$ = population mean breaking strength

Here for constructing 95% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

<u>So, 95% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

P(-2.02 < $t_4_4$ < 2.02) = 0.95 {As the critical value of t at 44 degree

of freedom are -2.02 & 2.02 with P = 2.5%}

P(-2.02 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.02) = 0.95

P( $-2.02 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.02 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.02 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.02 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

<u>95% confidence interval for</u> $\mu$ = [ $\bar X-2.02 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.02 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $768.2-2.02 \times {\frac{15.1}{\sqrt{45} } }$ , $768.2+2.02 \times {\frac{15.1}{\sqrt{45} } }$ ]

= [763.65 lb , 772.75 lb]

Therefore, 95% confidence interval for the mean breaking strength of the new steel cable is [763.65 lb , 772.75 lb].

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3 years ago

Does anyone know this? if so would you be kind to help me out?

Neko [114]

The answer is obtuse

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2 years ago

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for freshman year Nick had 3.5 for a gpa. what gpa does Nick needs for sophomore, junior, senior year to get the average of at l

JulsSmile [24]

Answer:4.0

Step-by-step explanation:

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2 years ago