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3 years ago

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The amount a spring will stretch varies directly with the amount of weight attached to the spring. If a spring stretches 2.1 inc

hes when 80 pounds is attached, how far will it stretch when 40 pounds is attached?

2 answers:

Alex73 [517]3 years ago

8 0

Answer:

A. 1.05 in.

Step-by-step explanation:

aleksklad [387]3 years ago

3 0

Answer:

$1.05\ in$

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $y/x=k$ or $y=kx$

Let

x -----> the amount of weight

y ----- the amount a spring will stretch

we know that

For x=80 lb, y=2.1 in

<em>Find the value of k (constant of proportionality)</em>

$k=y/x$

substitute the values

$k=2.1/80$

The linear equation is

$y=(2.1/80)x$

How far will it stretch when 40 pounds is attached?

For x=40 lb

substitute the value of x in the linear equation and solve for y

$y=(2.1/80)(40)=1.05\ in$

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Sloan [31]

Step-by-step explanation:

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you can check with the next problem to check if it is the same

25=k(20)

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Hope that helps :)

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3 years ago

If X = 10 cm, Y = 24 cm, and Z = 26 cm, what is the tangent of A?

omeli [17]

2.4cm is the tangent of A

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3 years ago

When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let

sergiy2304 [10]

Answer:

(a) P(X=3) = 0.093

(b) P(X≤3) = 0.966

(c) P(X≥4) = 0.034

(d) P(1≤X≤3) = 0.688

(e) The probability that none of the 25 boards is defective is 0.277.

(f) The expected value and standard deviation of X is 1.25 and 1.089 respectively.

Step-by-step explanation:

We are given that when circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%.

Let X = <em>the number of defective boards in a random sample of size, n = 25</em>

So, X ∼ Bin(25,0.05)

The probability distribution for the binomial distribution is given by;

$P(X=r)= \binom{n}{r} \times p^{r}\times (1-p)^{n-r} ; x = 0,1,2,......$

where, n = number of trials (samples) taken = 25

r = number of success

p = probability of success which in our question is percentage

of defectivs, i.e. 5%

(a) P(X = 3) = $\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$

= $2300 \times 0.05^{3}\times 0.95^{22}$

= <u>0.093</u>

(b) P(X $\leq$ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= $\binom{25}{0} \times 0.05^{0}\times (1-0.05)^{25-0}+\binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$

= $1 \times 1 \times 0.95^{25}+25 \times 0.05^{1}\times 0.95^{24}+300 \times 0.05^{2}\times 0.95^{23}+2300 \times 0.05^{3}\times 0.95^{22}$

= <u>0.966</u>

(c) P(X $\geq$ 4) = 1 - P(X < 4) = 1 - P(X $\leq$ 3)

= 1 - 0.966

= <u>0.034</u>

<u></u>

(d) P(1 ≤ X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3)

= $\binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$

= $25 \times 0.05^{1}\times 0.95^{24}+300 \times 0.05^{2}\times 0.95^{23}+2300 \times 0.05^{3}\times 0.95^{22}$

= <u>0.688</u>

(e) The probability that none of the 25 boards is defective is given by = P(X = 0)

P(X = 0) = $\binom{25}{0} \times 0.05^{0}\times (1-0.05)^{25-0}$

= $1 \times 1\times 0.95^{25}$

= <u>0.277</u>

(f) The expected value of X is given by;

E(X) = $n \times p$

= $25 \times 0.05$ = 1.25

The standard deviation of X is given by;

S.D.(X) = $\sqrt{n \times p \times (1-p)}$

= $\sqrt{25 \times 0.05 \times (1-0.05)}$

= <u>1.089</u>

8 0

3 years ago

A homemade loaf of bread turns out to be a perfect cube. Five slices of bread, each 0.6 inch thick, are cut from one end of the

erastovalidia [21]

A cube has all sides of equal length.

So let the side be x

that will be length = width = height = x

Now, when we cut the loaf into 5 equal slices, the length and height of the slices are the same as the original loaf. But the width will become less with each slicing. Here the width corresponds to the thickness of the slices.

For each slice the dimensions are -

Length = x

height = x

width = 0.6

Volume of the original loaf is x³.

Volume of 5 slices is 5(x * x * 0.6).

Volume that is left after the cutting the slices is 235 inch³.

So, equation of the volume is=

x³ - 5(0.6x²) = 235

Solve for x using this equation.

x³ - 3x² = 235

x³ - 3x² - 235 = 0

Using a graph calculator to find the roots, we get

x = 7.35

Hence the dimensions of the original loaf is 7.35 inches for each side.

Please find attached the graph.

7 0

3 years ago

Write three expressions that are equivalent to 8x + 16.

anyanavicka [17]

Answer:

1 is 8x + 16 = 6x

Step-by-step explanation:

4 0

3 years ago