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sertanlavr [38]
3 years ago
9

Need ASAP!! Plsss I’ll give brainliest!! Do the first and second pls ):

Mathematics
1 answer:
Genrish500 [490]3 years ago
4 0

Answer:

Q1

14+5+11=30

360÷30=12

14×12=168,5×12=60,11×12=132

Step-by-step explanation:

Q2

9+7+3=19

266÷19=14

shortest side is the lowest ratio that is 3

so 3×14=42

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Step-by-step explanation:

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al receives $25 weekly allowance. he also babysits his two younger siblings on weekends and is paid $9 per hour. al's friend, Ma
Harlamova29_29 [7]
To answer this question you would set this up as an equation with all the information about Al on one side being equal to all the information for Matthew on the other side. This is what it would look like: $25 + $9h= $14h. h represents the number of hours they work. It they each worked 5 hours a week, they would make the same amount of money. I determined this by solving the equation for h. I subtracted 9h on both sides leaving you with $25=5h. H would equal 5 hours.
7 0
3 years ago
Assume that when adults with smartphones are randomly​ selected, 51​% use them in meetings or classes. If 11 adult smartphone us
Luba_88 [7]

Answer:

The probability is 0.2356.

Step-by-step explanation:

Let X is the event of using the smartphone in meetings or classes,

Given,

The probability of using the smartphone in meetings or classes, p = 51 % = 0.51,

So, the probability of not using smartphone in meetings or classes, q = 1 - p = 1 - 0.51 = 0.49,

Thus, the probability that fewer than 5 of them use their smartphones in meetings or classes.

P(X<5) = P(X=0) + P(X=1) + P(X=2) + P(X=3)+P(X=4)

Since, binomial distribution formula is,

P(x) = ^nC_r p^x q^{n-x}

Where, ^nC_r=\frac{n!}{r!(n-r)!}

Here, n = 11,

Hence,  the probability that fewer than 5 of them use their smartphones in meetings or classes

=^{11}C_0 (0.5)^0 0.49^{11}+^{11}C_1 (0.5)^1 0.49^{10}+^{11}C_2 (0.5)^2 0.49^{9}+^{11}C_3 (0.5)^3 0.49^{8}+^{11}C_4 (0.5)^4 0.49^{7}

=(0.5)^0 0.49^{11}+11(0.5)0.49^{10} + 55(0.5)^2 0.49^{9}+165 (0.5)^3 0.49^{8} +330(0.5)^4 0.49^{7}

=0.235596671797

\approx 0.2356

4 0
4 years ago
Could 1 2 . 2 cm , 6 . 0 cm , 12.2 cm,6.0 cm, and 4 . 2 cm 4.2 cm be the side lengths of a triangle?
HACTEHA [7]

Answer:

Yes they can be


8 0
3 years ago
Y’all please help me
OLga [1]
I think it is 2 but I really don’t know
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