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2 years ago

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A metal disc of 12cm in diameter and 5cm thick is melted down and cast into a cylindrical bar of diameter 5cm .How long is the b

ar

1 answer:

Trava [24]2 years ago

3 0

Answer:

d bar

Step-by-step explanation:

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At an intersection, the red-light times are normally distributed with a mean time of 3 minutes and a standard deviation of 0.25

denpristay [2]

Percent of red lights last between 2.5 and 3.5 minutes is 95.44% .

<u>Step-by-step explanation:</u>

Step 1: Sketch the curve.

The probability that 2.5<X<3.5 is equal to the blue area under the curve.

Step 2:

Since μ=3 and σ=0.25 we have:

P ( 2.5 < X < 3.5 ) =P ( 2.5−3 < X−μ < 3.5−3 )

⇒ P ( (2.5−3)/0.25 < (X−μ)/σ < (3.5−3)/0.25)

Since, Z = (x−μ)/σ , (2.5−3)/0.25 = −2 and (3.5−3)/0.25 = 2 we have:

P ( 2.5<X<3.5 )=P ( −2<Z<2 )

Step 3: Use the standard normal table to conclude that:

P ( −2<Z<2 )=0.9544

Percent of red lights last between 2.5 and 3.5 minutes is $0.9544(100) = 95.44$ % .

3 0

2 years ago

The coordinates of the vertices of △PQR are P(1, 4) , Q(2, 2) , and R(−2, 1) . The coordinates of the vertices of △P′Q′R′ are P′

sveticcg [70]

ANSWER

The correct answer is B

<u>EXPLANATION</u>

If we analyse the coordinates carefully, you realize there is a mapping of

$(x,y)\rightarrow(-x,y)$

This is a reflection in the y-axis.

Since the transformation is a reflection, the shape is preserved. Therefore, the image triangle P'Q'R' of triangle

PQR are congruent.

5 0

2 years ago

Read 2 more answers

One of the factors of g2 − 28g + 196 is a. g-7 b. g-14 c. g+14 d. g+7

EastWind [94]

$g^{2}-28g+196=g^{2}-14g-14g+196=g(g-14)-14(g-14)= \\ \\ (g-14)(g-14)=(g-14)^{2}$

P.S :>

$(a-b)^{2}=a^{2}-2ab+b^{2}$

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4 0

2 years ago

Read 2 more answers

A producer of steel cables wants to know whether the steel cables it produces have an average breaking strength of 5000 pounds.

Sever21 [200]

Answer:

Step-by-step explanation:

Hello!

The producer needs his cables to have an average breaking strength of 5000 pounds. A lower average breaking strength means that the cable is not adequate, a higher average breaking strength results in an unnecessary increase in production costs.

The sample is of 64 steel cables and the breaking strength of the pieces was recorded.

I always recommend that the first step to any statistic exercise is to establish the study variable that way you'll have fresh in mind the kind of data set you are working with and the type of distribution to expect from them (for example if it is a discrete variable you'd expect a binomial distribution, a continuous variable leads to a normal distribution (exact or approximate) a categorical variable sets the path to work with non-parametrical statistics such as Chi-Square statistics.)

In this example the study variable is:

X: Breaking strength of a steel cable.

This variable is continuous so first I'll use the sample information to test its distribution. Keep in mind that one of the conditions to use the Student t-test is that the variable has a normal distribution.

The p-value for the normality test is 0.8512, comparing it with the level of significance of the test (α: 0.05) the decision is to not reject the null hypothesis of the normality test, so you can conclude that the breaking strength of the steel cables has a normal distribution:

X~N(μ;σ²)

1.

The standard error of the test is the square root of the variance.

Using the following formula you have to calculate the variance:

$S^2= \frac{1}{n-1}*(sumX^2-(\frac{(sumX)^2}{n} ))$

n=64

∑X= 330174.88

∑X²= 1718980202.34

$S^2= \frac{1}{63}*(1718980202.34-(\frac{(330174.88)^2}{64} ))$

S²= 2478376895

S= 497.8329 ≅497.833

2.

For this test the hypotheses are:

H₀: μ = 5000

H₁: μ ≠ 5000

$t= \frac{Xbar-Mu}{\frac{S}{\sqrt{n} } }$

The sample mean is:

Xbar= ∑X/n)= 330174.88/64=5158.98

‬ $t= \frac{5158.98-5000}{\frac{497.833}{\sqrt{64} } }$

t= 158.98/62.229= 2.555

3.

This test is two-tailed and so is the p-value. I've used statistics software to calculate it:

p-value 0.0131

4.

Using a significance level of 5%, since the p-value is less than α, the decision is to reject the null hypothesis. You can conclude at this level that the average breaking strength of the steel cables is different than 5000.

I hope this helps!

6 0

2 years ago

A circle with radius of \greenD{6\,\text{cm}}6cmstart color #1fab54, 6, start text, c, m, end text, end color #1fab54 sits insid

Diano4ka-milaya [45]

Answer: $141 cm^{2}$

Step-by-step explanation:

We have two circles:

Cirlce 1, with a radius $r=6 cm$ and area $A_{1}$ :

$A_{1}=\pi r^{2}$

And Cicle 2, with a radius $R=9 cm$ and area $A_{2}$ :

$A_{1}=\pi R^{2}$

Since Circle 1 is inside Circle 2 and assuming the area of the shaded region is the shown in the attached image, its area is:

$A=A_{2}-A_{1}=\pi R^{2}-\pi r^{2}$

$A=\pi (R^{2}- r^{2})$

$A=\pi ((9cm)^{2}- (6cm)^{2})$

Finally:

$A=141.37 cm^{2} \approx 141 cm^{2}$

3 0

2 years ago