Write the coordinates of the vertices after a reflection across the line y = x.

A tea bag is shaped like a square pyramid with the base measuring 4 centimeters on each side. The dlant height is 4.5 centimeter

Lynna [10]

A square pyramid is a pyramid that has a square as its base.Thus, its sides are composed of 4 triangles.

We need to get the area of the square and the triangles to get the total surface area.

Square sides = 4 cm.
triangle: base = 4cm ; height = 4.5

Area of the square = s² = (4cm)² = 16 cm²
Area of the triangle = hb/2 = 4.5cm(4cm) /2 = 9 cm²
9 cm² x 4 triangles = 36 cm²

Total area = 16cm² + 36cm² = 52 cm² is the area of the required mesh to create the tea bag.

6 0

3 years ago

Arrange the following in ascending order 3/4 , -1/3 , 1/-2 , 3/10

Over [174]

Answer:

Ans: 1/-2 , -1/3 , 3/10 , 3/4

Step-by-step explanation:

The LCM of 2, 3, 4 and 10 is 60.

Let's make denominator equal,

→ 3/4 , -1/3 , 1/-2 , 3/10

→ 3×15/4×15 , -1×20/3×20 , -1×30/2×30 , 3×6/10×6

→ 45/60 , -20/60 , -30/60 , 18/60

Now we can order in ascending,

→ -30/60 , -20/60 , 18/60 , 45/60

→ 1/-2 , -1/3 , 3/10 , 3/4 {final answer}

7 0

3 years ago

A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as

Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

Step-by-step explanation:

Let X = number of seconds in the batch.

The probability of the random variable X is, p = 0.31.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X is:

$P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...$

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

$P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888$

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

$=1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776$

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

$={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453$