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3 years ago

How many lines of symmetry does a kite have? 4 3 1 2

Mathematics

1 answer:

skelet666 [1.2K]3 years ago

4 0

A typical quadrilateral kite (kinda looks like a diamond) only has 1 line of symmetry

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Trudy works at a job centre.

denpristay [2]

<span>To solve the problem, to 40% of the total customer should be solve. And this can be solve by muliplting it by the fraction (0.4). so 140 x 0.4 = 56 customers. Since Trudy has already 60 of her sutomers job ready then she met her target.</span>

3 0

3 years ago

Vince added 0.912,0.882 ,and 1.354 liters of liquid to a bottle.About how much is in the bottle?

Levart [38]

3.148 liters of liquid.

3 0

3 years ago

What constant acceleration is needed to accelerate a baseball from 6 feet/sec to 116 feet/sec in 2 seconds?

Vesnalui [34]

Answer:

d. 55 feet /sec^2

Step-by-step explanation:

Average acceleration is ...

a = ∆v/∆t = (116 ft/s -6 ft/s)/(2 s)

= 110/2 ft/s^2 = 55 ft/s^2

4 0

2 years ago

Which expression is equivalent to2/3-1/4x+1/2x-1/6?

ivann1987 [24]

2/3 - 1/4 ... let's find the common numbers 12 so new numbers would be 8/12 and 3/12 aka 11/12 + .. anyways it's 11/12 plus 2/3 x

6 0

2 years ago

In a particular course, it was determined that only 70% of the students attend class on Fridays. From past data it was noted tha

siniylev [52]

Answer: Probability that students who did not attend the class on Fridays given that they passed the course is 0.043.

Step-by-step explanation:

Since we have given that

Probability that students attend class on Fridays = 70% = 0.7

Probability that who went to class on Fridays would pass the course = 95% = 0.95

Probability that who did not go to class on Fridays would passed the course = 10% = 0.10

Let A be the event students passed the course.

Let E be the event that students attend the class on Fridays.

Let F be the event that students who did not attend the class on Fridays.

Here, P(E) = 0.70 and P(F) = 1-0.70 = 0.30

P(A|E) = 0.95, P(A|F) = 0.10

We need to find the probability that they did not attend on Fridays.

We would use "Bayes theorem":

$P(F\mid A)=\dfrac{P(F).P(A\mid F)}{P(E).P(A\mid E)+P(F).P(A\mid F)}\\\\P(F\mid A)=\dfrac{0.30\times 0.10}{0.70\times 0.95+0.30\times 0.10}\\\\P(F\mid A)=\dfrac{0.03}{0.695}=0.043$

Hence, probability that students who did not attend the class on Fridays given that they passed the course is 0.043.

8 0

3 years ago