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bazaltina [42]
3 years ago
7

If you know how to solve this, Please answer it. Thank You

Mathematics
1 answer:
spayn [35]3 years ago
8 0

Answer:

48

Step-by-step explanation:

x1 = 20 ; y1 = 12  

x2 = ? ; y2 = 5

\frac{x_{1}}{x_{2}}=\frac{y_{2}}{y_{1}}\\\\\frac{20}{x_{2}}=\frac{5}{12}\\\\20 *12 = x_{2}*5\\\\\frac{20*12}{5}=x_{2}\\\\4*12=x_{2}\\\\x_{2}=48

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You can buy 3 pizzas and 3 drinks
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3 years ago
Use an inequality symbol to compare -3 +7____-10-2
Damm [24]

Answer:

-3+7 > -10-2

Step-by-step explanation:

-3+7 = +4

-10-2 = -12

∴ +4 > -12

7 0
3 years ago
In a​ study, 36​% of adults questioned reported that their health was excellent. A researcher wishes to study the health of peop
steposvetlana [31]

Answer:

0.3907

Step-by-step explanation:

We are given that 36​% of adults questioned reported that their health was excellent.

Probability of good health = 0.36

Among 11 adults randomly selected from this​ area, only 3 reported that their health was excellent.

Now we are supposed to find the probability that when 11 adults are randomly​ selected, 3 or fewer are in excellent health.

i.e. P(x\leq 3)=P(x=1)+{P(x=2)+P(x=3)

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

p is the probability of success i.e. p = 0.36

q = probability of failure = 1- 0.36 = 0.64

n = 11

So, P(x\leq 3)=P(x=1)+{P(x=2)+P(x=3)

P(x\leq 3)=^{11}C_1 (0.36)^1 (0.64)^{11-1}+^{11}C_2 (0.36)^2 (0.64)^{11-2}+^{11}C_3 (0.36)^3 (0.64)^{11-3}

P(x\leq 3)=\frac{11!}{1!(11-1)!} (0.36)^1 (0.64)^{11-1}+\frac{11!}{2!(11-2)!}  (0.36)^2 (0.64)^{11-2}+\frac{11!}{3!(11-3)!} (0.36)^3 (0.64)^{11-3}

P(x\leq 3)=0.390748

Hence  the probability that when 11 adults are randomly​ selected, 3 or fewer are in excellent health is 0.3907

5 0
3 years ago
Can someone please help
Ira Lisetskai [31]
E = 1/2(1.59 × 10^3)([2.7 × 10^1]^2)
plug it in to the o'l calculator and ¡bam!
4 0
3 years ago
Plot the line for the equation on the graph.<br><br><br><br><br><br><br> y−4=−14(x−5)
nevsk [136]
Y=mx+b

Y= -14(x-5)+4
Y= -14x+70+4
Y= -14x+74

Plot (0,74)
Then move down 14 then over left 1
8 0
3 years ago
Read 2 more answers
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