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2 years ago

5

Suppose you have a rectangular shipping box that measures 30 cm × 15 cm × 20 cm. What is the longest length of a 1 cm diameter w

ooden dowel that could fit inside the box? (round to the nearest whole centimeter)
PLEASE HURRY

2 answers:

balandron [24]2 years ago

5 0

the box is cuboid in shape.

Longest wooden dowel that could fit the box is equal to length of diagonal of cuboid

length of diagonal of cuboid = sqrt(l^2+b^2+h^2)= sqrt(1525)

ngest wooden dowel that could fit the box is equal = 39.05 = 39 cm

mr_godi [17]2 years ago

3 0

See what happened was you failed .

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If pentagon ABCDE is rotated 180 degrees around the origin to create pentagon A’B’C’D’E’, what is the ordered pair of point A’?

Semmy [17]

Answer:

(A)

Step-by-step explanation:

Point A is at (-2,4)

Rule of a 180° rotation about the origin: (x,y) --> (-x,-y)

Using the rule, (-2,4) will become (2,-4).

A' should be (2,-4) or Option A.

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2 years ago

Triangles E F G and E prime F prime G prime are shown. Angles F E G and F prime E prime G prime are 72 degrees. Angles E F G and

Karolina [17]

Answer:

We need a non-included side of one triangle

Step-by-step explanation:

By means of the AAS postulate.

The Angle-Angle-Side postulate (AAS) tells us that if two angles and a non-included side of one triangle are congruent to two angles and the corresponding non-included side of another triangle, then the two triangles are congruent.

3 0

2 years ago

The Acme Candy Company claims that 60% of the jawbreakers it produces weigh more than 0.4 ounces. Suppose that 800 jawbreakers

vichka [17]

Answer:

Yes, it would be statistically significant

Step-by-step explanation:

The information given are;

The percentage of jawbreakers it produces that weigh more than 0.4 ounces = 60%

Number of jawbreakers in the sample, n = 800

The mean proportion of jawbreakers that weigh more than 0.4 = 60% = 0.6 = $\mu _ {\hat p}$ =p

The formula for the standard deviation of a proportion is $\sigma _{\hat p} =\sqrt{\dfrac{p(1-p)}{n} }$

Solving for the standard deviation gives;

$\sigma _{\hat p} =\sqrt{\dfrac{0.6 \cdot (1-0.6)}{800} } = 0.0173$

Given that the mean proportion is 0.6, the expected value of jawbreakers that weigh more than 0.4 in the sample of 800 = 800*0.6 = 480

For statistical significance the difference from the mean = 2× $\sigma _{\hat p}$ = 2*0.0173 = 0.0346 the equivalent number of Jaw breakers = 800*0.0346 = 27.7

The z-score of 494 jawbreakers is given as follows;

$Z=\dfrac{x-\mu _{\hat p} }{\sigma _{\hat p} }$

$Z=\dfrac{494-480 }{0.0173 } = 230.94$

Therefore, the z-score more than 2 × $\sigma _{\hat p}$ which is significant.

8 0

2 years ago

The prices of three t-shirts styles are $24, $30 and $36. the probability of choosing a $24 t-shirt is 1/6. the probability of c

Slav-nsk [51]

$\text{Answer} : \text{The expected value of a t-shirt is \$31.}$

Explanation:

Since we have given that

The prices of three t-shirts styles i.e $24, $30, $36 with their probability is given by

$\frac{1}{6}, \frac{1}{2},\frac{1}{3}$

As we know that,

$E(X)= \sum_{1}^{3}x_iP(x_i)$

$\text{where} x_i \text{ is the prices of t- shirts styles}$

Now,

$x_1= \$24 , x_2=\$30 , x_3=$36$

and

$P(x_1)=\frac{1}{6},P(x_2)=\frac{1}{2}, P(x_3)=\frac{1}{3}$

So,

$E(X)= 24\times \frac{1}{6}+30\times\frac{1}{2}+36\times \frac{1}{3}\\=4+15+12\\=31$

So, the expected value of a t-shirt = $31.

4 0

2 years ago

A textbook has 500 pages on which typographical errors could occur. Suppose that there are exactly 10 such errors randomly locat

DiKsa [7]

Answer:

The probability of a selection of 50 pages will contain no errors is 0.368

The probability that the selection of the random pages will contain at least two errors is 0.2644

Step-by-step explanation:

From the information given:

Let q represent the no of typographical errors.

Suppose that there are exactly 10 such errors randomly located on a textbook of 500 pages. Let $\mu$ be the random variable that follows a Poisson distribution, then mean $\mu = \dfrac{10}{500}= 0.02$

and the mean that the random selection of 50 pages will contain no error is $\lambda = 50 \times 0.02 =1$

∴

$Pr(q= 0) = \dfrac{e^{-1} (1)^0}{0!}$

Pr(q =0) = 0.368

The probability of a selection of 50 pages will contain no errors is 0.368

The probability that 50 randomly page contains at least 2 errors is computed as follows:

P(X ≥ 2) = 1 - P( X < 2)

P(X ≥ 2) = 1 - [ P(X = 0) + P (X =1 )] since it is less than 2

$P(X \geq 2) = 1 - [ \dfrac{e^{-1} 1^0}{0!} +\dfrac{e^{-1} 1^1}{1!} ]$

$P(X \geq 2) = 1 - [0.3678 +0.3678]$

$P(X \geq 2) = 1 -0.7356$

P(X ≥ 2) = 0.2644

The probability that the selection of the random pages will contain at least two errors is 0.2644

6 0

2 years ago