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3 years ago

If 8 girls brought a lunch box and 6 boys didnt what is the probability of a student bringing one.

Mathematics

1 answer:

Tatiana [17]3 years ago

3 0

Answer:

43 percent

Step-by-step explanation:

6/14 ×100%

=0.248

=43%

first add up the total people, then substract the total of people to the people that didn't bring them, divide them then multiply it by 100%

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3 geometry questions, 30 points!

polet [3.4K]

Answer:

A ) The value of x for the given circle with chords and center is 8.3

B) The circumference of circle with chords 1.2 cm and 0.5 cm is 4.082

Step-by-step explanation:

Given two figures :

<u>For Figure first </u>

A circle with center y , having two chords FM and NM

FM = 5 x

MN = 2 x + 25

Now from theorem of circle ,

Chords equidistant from center of circle are equal in length

I.e distance of chord MN from center y and distance of FM from center y are equal

So, FM = MN

Or, 5 x = 2 x + 25

Or, 5 x - 2 x = 25

Or, 3 x = 25

∴ x = $\frac{25}{3}$ = 8.33

<u>For figure second</u>

The length of two adjacent chords of circle is 1.2 cm and 0.5 cm

Let the center of circle = O

Length of chord AB = 1.2 cm

Length of chord BC = 0.5 cm

As both chords are at 90° to each other

So The Length of diameter of circle AC = $\sqrt{AB^{2}+BC^{2}}$

Or, The Length of diameter of circle AC = $\sqrt{1.2^{2}+0.5^{2}}$

Or, The Length of diameter of circle AC = $\sqrt{1.44+0.25}}$

Or, The Length of diameter of circle AC = $\sqrt{1.69}$

∴ The Length of diameter of circle AC = 1.3 cm

So, Circumference of circle = $\pi d$

Or, Circumference of circle = 3.14 × 1.3

∴ Circumference of circle = 4.082 cm

Hence,

A ) The value of x for the given circle with chords and center is 8.3

B) The circumference of circle with chords 1.2 cm and 0.5 cm is 4.082 Answer

8 0

3 years ago

Read 2 more answers

Calculate the area and volume of 3.3 m and 6.2 m

Andre45 [30]

Answer:

Area is 20.46

Step-by-step explanation:

Area is length × breadth

6.2 × 3.3 = 20.46

The volume can't be found since there is no height

PLEASE MARK BRAINLIEST

8 0

2 years ago

Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution. At Meadowbrook Hospital, the mean we

Marina86 [1]

Answer:

Required Probability = 0.1283 .

Step-by-step explanation:

We are given that at Meadow brook Hospital, the mean weight of babies born to full-term pregnancies is 7 lbs with a standard deviation of 14 oz.

Firstly, standard deviation in lbs = 14 ÷ 16 = 0.875 lbs.

Also, Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution.

Let X = mean weight of the babies, so X ~ N( $\mu = 7 lbs , \sigma^{2} = 0.875^{2} lbs$ )

The standard normal z distribution is given by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, X bar = sample mean weight

n = sample size = 4

Now, probability that the average weight of the four babies will be more than 7.5 lbs = P(X bar > 7.5 lbs)

P(X bar > 7.5) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{7.5-7}{\frac{0.875}{\sqrt{4} } }$ ) = P(Z > 1.1428) = 0.1283 (using z% table)

Therefore, the probability that the average weight of the four babies will be more than 7.5 lbs is 0.1283 .

8 0

3 years ago

31. I just need to do it

NikAS [45]

You need to buy 6 notesbooks

5 0

3 years ago

For this lease plan, find the total cost for the month.

Levart [38]

Answer:

C. 873.67

Step-by-step explanation:

399.17 + 150 + 69 + 200 + 55.5 = 873.67

8 0

3 years ago