Which statement is true about a line and a point? (1 point)

You spend $11 on school supplies. You purchase pencils of $0.25 each and pens for $2 each. You purchase a total of 30 pencils an

Klio2033 [76]

28 pencils and 2 pens

5 0

2 years ago

n ΔXYZ, m∠X = 90° and m∠Y = 30°. In ΔTUV, m∠U = 30° and m∠V = 60°. Which is true about the two triangles? ΔXYZ ≅ ΔTUV ΔXYZ ≅ ΔVU

lions [1.4K]

The triangles are similar because corresponding angles are congruent, however no congruency statement can be made because the side lengths are unknown.

The 4th selection is appropriate.

5 0

2 years ago

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A building has an entry the shape of a parabolic arch 74 ft high and 28 ft wide at the base as shown below.

sergiy2304 [10]

The equation of a parabola with vertex at (h,k) is
y=a(x-h)²+k

vertex isi at (0,0)

y=a(x-0)²+0
y=a(x)²
y=ax²
find a

we see that one point is (14,-74)
x=14 and y=-74

-74=a(14²)
-74=196a
divide both sides by 196
-37/98=a

the equation is

$y= \frac{-37}{98} x^2$

6 0

2 years ago

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The manager at Gabriela's Furniture Store is trying to figure out how much to charge for a book shelf that just arrived. The boo

maria [59]

The shelf should sell for $235.20.

Marking the price up by 60% means taking 160% of the cost:
160% = 160/100 = 1.6; 1.6(147) = 235.20

5 0

2 years ago

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In choice situations of this type, subjects often exhibit the "center stage effect," which is a tendency to choose the item in t

victus00 [196]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The probability that he or she would choose the pair of socks in the center position is $p =\frac{1}{5}$

The correct answer choice is

X has a binomial distribution with parameters n=100 and p=1/5

b

The mean is $\mu = 20$

The standard deviation is $\sigma=4$

c

The probability, $P =0.0002$

d

The correct answer is

The experiment supports the center stage effect. If participants were truly picking the socks at random, it would be highly unlikely for 34 or more to choose the center pair.

Using the R the probability $Pe = 0.0003$

The probabilities $P \approx Pe$

Step-by-step explanation:

Since the person selects his or her desired pair of socks at random , then the probability that the person would choose the pair of socks in the center position from all the five identical pair is mathematically evaluated as

$p =\frac{1}{5}$

$=0.2$

The mean of this distribution is mathematical represented as

$\mu = np$

substituting the value

$\mu = 100 * 0.2$

$\mu = 20$

The standard deviation is mathematically represented as

$\sigma = \sqrt{np (1-p)}$

substituting the value

$= \sqrt{100 * 0,2 (1-0.2)}$

$\sigma=4$

Applying normal approximation the probability that 34 or more subjects would choose the item in the center if each subject were selecting his or her preferred pair of socks at random would be mathematically represented as

$P=P(X \ge 34 )$

By standardizing the normal approximation we have that

$P(X \ge 34) \approx P(Z \ge z)$

Now z is mathematically evaluated as

$z = \frac{x-\mu}{\sigma }$

Substituting values

$z = \frac{34-20}{4}$

$=3.5$

So using the z table the $P(Z \ge 3.5)$ is 0.0002

The probability P and Pe that 34 or more subject would choose the center pair is very small So

The correct answer is

The experiment supports the center stage effect. If participants were truly picking the socks at random, it would be highly unlikely for 34 or more to choose the center pair.

6 0

2 years ago