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Dmitry [639]
2 years ago
8

15 + 45 = 5( blank + 9)

Mathematics
2 answers:
Tatiana [17]2 years ago
5 0

Answer:

Step-by-step explanation:

15 + 45 = 5( _ + 9)

60 = 5 ( 3 + 9)

60 = 5(12)

60 = 60

Ludmilka [50]2 years ago
5 0

Answer:

3

Step-by-step explanation:

5*12=60

9+3=12

5*12=60

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Apollo Spas services 156 hot tubs. If each hot tub needs 105 mL of muriatic acid, how many liters of acid are needed for all of
True [87]

Answer:

16380 mL or 16.380 L

Step-by-step explanation:

1 hot tub requires 105 mL of Acid

156 hot tubs requires 105 * 156 mL of acid = 16380 mL of acid are needed.

16380 / 1000 mL/L = 16.380 Liters.

5 0
2 years ago
A scientist inoculates​ mice, one at a​ time, with a disease germ until he finds 3 that have contracted the disease. If the prob
UkoKoshka [18]

Answer:

The probability that 8 mice are​ required is 0.2428.

Step-by-step explanation:

Given : A scientist inoculates​ mice, one at a​ time, with a disease germ until he finds 3 that have contracted the disease. If the probability of contracting the disease is two sevenths.

To find : What is the probability that 8 mice are​ required? The probability that that 8 mice are required is nothing ?

Solution :

Applying binomial distribution,

P(X=r)=^nC_r p^rq^{n-r}

Where, p is the probability of success p=\frac{2}{7}

q is the probability of failure q=1-p, q=1-\frac{2}{7}=\frac{5}{7}

n is total number of trials n=8

r=3

Substitute the values,

P(X=3)=^8C_3 (\frac{2}{7})^3 (\frac{5}{7})^{8-3}

P(X=3)=\frac{8!}{3!5!}\times \frac{8}{343}\times (\frac{5}{7})^{5}

P(X=3)=\frac{8\times 7\times 6}{3\times 2\times 1}\times \frac{8}{343}\times \frac{3125}{16807}

P(X=3)=0.2428

Therefore, the probability that 8 mice are​ required is 0.2428.

5 0
3 years ago
There were 16 sandwiches on a tray for a lunch meeting. After the meeting, 3 were left. How many sandwiches were eaten?
kifflom [539]

Answer:

13

Step-by-step explanation:

16-3+= 13

6 0
3 years ago
Thank you in advance for the help !!
babymother [125]

The Answer is C, my good friend.

4 0
2 years ago
Find the zeros of x3 +2X^2-23x-60
laila [671]
Hello,

P(x)=x^3+2x²-23x-60
P(-4)=(-4)^3+2*(-4)²-23*(-4)-60=0
P(-3)=(-3)^4+2*(-3)²-23*(-3)-60=0
P(5)=5^3+2*5²-23*5-60=0
Zeros are -4,-3,5


5 0
2 years ago
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