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3 years ago

Please help me on this one

Mathematics

2 answers:

Annette [7]3 years ago

3 0

Divide $435.15 by 9 = $48.35 unit price

Vlad1618 [11]3 years ago

3 0

$48.35? I divided $435.15 divided by 9

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Insert a digit to make the number divisible by 24 if possible: 44_8 pls help

Nat2105 [25]

Answer:

it's 4488 Wich would make 187

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The base of a triangle is 3 cm longer than its altitude. The area of the triangle is 35cm^2. Find the altitude. (Hint: use the a

klemol [59]

The base is 3+(x), altitude is x so substitute. Now we know the area of a triangle is base X height X 1/2. Substitute again! 1/2 (3+x)(x)=35. Multiply both sides by 2 to cancel out the 1/2. Now you have (x)(x+3)=70 and you have to foil out the left side x^2+3x=70. Subtract 70 on both sides x^2+3x-70=0. Find two numbers that multiply to -70 and add to 3. Solve (x+10)(x-7)=0. the x value is 7. since you can't have negative length values. Substitute 7 into 3+x for the base so you know the base is 10 and the height is 7.

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3 years ago

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.

So, 95% confidence interval for the population mean, $\mu$ is ;

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

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

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

= [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].

3 0

3 years ago