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Rom4ik [11]
3 years ago
5

Micaela is 2 years older than Sam. In 4 years, the sum of their ages will be 40. How old is Micaela now?

Mathematics
1 answer:
ohaa [14]3 years ago
6 0
If Micael is 2 years older than sam and sam is 15 than than micael is 17. 
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Allison is rolling her hula hoop on the playground. The radius of her hula hoop is 35 cm. What is the distance the hula hoop rol
lidiya [134]

Answer:

Distance covered by hula hoop rolls in 4 full rotations is 880 cm .

Step-by-step explanation:

Formula

Perimeter\ of\ a\ circle = 2\pi r

Where r is the radius of the circle.

As given

Allison is rolling her hula hoop on the playground.

The radius of her hula hoop is 35 cm.

r = 35 cm

\pi = \frac{22}{7}

Putting the value in the formula

Perimeter\ of\ a\ hula hoop = 2\times \frac{22}{7} \times 35

                                             = 220 cm

As given

The hula hoop rolls in 4 full rotations.

Distance covered by hula hoop rolls in 4 full rotations =  220 × 4

                                                                                          = 880 cm

Therefore the Distance covered by hula hoop rolls in 4 full rotations is 880 cm .

6 0
3 years ago
Which fractions are equivalent to 2/8?
Hitman42 [59]

Answer:

1/4 and 3/12

Step-by-step explanation:


3 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B7%7D%7B4%7D%20%3D%20%5Cfrac%7B21%7D%7B14%7D%20" id="TexFormula1" title=" \frac{7}
aev [14]
This looks super duper hard wish I can solve it out
8 0
3 years ago
In the data set below, what is the interquartile range?
satela [25.4K]

Answer:

53

Step-by-step explanation:

To find the interquartile range (IQR), ​first find the median (middle value) of the lower and upper half of the data. These values are quartile 1 (Q1) and quartile 3 (Q3). The IQR is the difference between Q3 and Q1.

3 0
2 years ago
32​% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a
Tems11 [23]

Answer:

a) P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

b) P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668

c) P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816

Step-by-step explanation:

1) Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

2) Solution to the problem

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=10, p=0.32)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

Part a

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

Part b

P(X> 2)=1-P(X\leq 2)=1-[P(X=0)+P(X=1)+P(X=2)]

P(X=0)=(10C0)(0.32)^0 (1-0.32)^{10-0}=0.0211  

P(X=1)=(10C1)(0.32)^1 (1-0.32)^{10-1}=0.0995

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668

Part c

P(2 \leq x \leq 5)=P(X=2)+P(X=3)+P(X=4)+P(X=5)

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

P(X=3)=(10C3)(0.32)^3 (1-0.32)^{10-3}=0.264

P(X=4)=(10C4)(0.32)^4 (1-0.32)^{10-4}=0.218

P(X=5)=(10C5)(0.32)^5 (1-0.32)^{10-5}=0.123

P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816

6 0
3 years ago
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