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4 years ago

Mr. Beeson has $220 in the bank and $12 in his billfold. He has bills of $76 and $188 to pay. Use a

Mathematics

1 answer:

eimsori [14]4 years ago

4 0

Answer:

Your answer is -32

Step-by-step explanation:

First you add $220 and $12 together giving you $232

Then you subtract $76 from the $232 giving you $156

Then subtract $188 from $156 giving you a total of -$-32

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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.

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

4 years ago

What is the vertex , x coordinates , and x intercepts of y=2x ^2 + 4x-30

Evgesh-ka [11]

5.1 Find the Vertex of y = x2-2x-15

Parabolas have a highest or a lowest point called the Vertex . Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) . We know this even before plotting "y" because the coefficient of the first term, 1 , is positive (greater than zero).

Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.

Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.

For any parabola,Ax2+Bx+C,the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is 1.0000 

Plugging into the parabola formula 1.0000 for x we can calculate the y -coordinate :
 y = 1.0 * 1.00 * 1.00 - 2.0 * 1.00 - 15.0
 or y = -16.000

6 0

3 years ago

Question 1 of 7

vagabundo [1.1K]

Formula:
$rs\pi + {r}^{2} \pi$
π(6)(8) + π(6)(6)

48π + 36π

84π

84( 3.14 )

Answer: 263.76 cm^2

IGNORE, IT IS WRONG

7 0

3 years ago

what are the values of a, b, and c in the quadratic equation 0 = x2 – 3x – 2? - a = , b = 3, c = 2 - a = , b = –3, c = –2 - a =

Dafna11 [192]

$x^2-3x-2=0\\\\a=1;\ b=-3;\ c=-2\\\Delta=b^2-4ac\\\\\Delta=(-3)^2-4\cdot1\cdot(-2)=9+8=17 \ \textgreater \ 0\\\\then\\x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\x_1=\dfrac{3-\sqrt{17}}{2\cdot1}=\boxed{\dfrac{3-\sqrt{17}}{2}}\\\\x_2=\dfrac{3+\sqrt{17}}{2\cdot1}=\boxed{\dfrac{3+\sqrt{17}}{2}}$