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4 years ago

I WILL MARK BRAINIEST IF CORRECT!

Mathematics

2 answers:

Lunna [17]4 years ago

8 0

Answer:

I believe I would be the last one D

alex41 [277]4 years ago

5 0

Answer:

Last choice

Step-by-step explanation:

He spins the spinner and can get Red, White, or Blue.

For each of those three outcomes, the coin can land on Heads or Tails.

The sample space has the following 6 possible outcomes.

Red, Heads

Red, Tails

White, Heads

White, Tails

Blue, Heads

Blue, Tails

The answer is the last choice.

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Submit your program using the zyBooks system by March 15, 2020, 11:59 pm. Please submit original work. Implement the following c

Klio2033 [76]

Answer:

def validateID(this_id):

is_valid = 0

checksum = 0

n = int(this_id)

count=1

while(n!=0):

d=int(n%10)

if(count%2==0):

d=d*2

if(d<10):

checksum=checksum+d

else:

while(d!=0):

x = int(d%10)

checksum = checksum + x

d=int(d/10)

else:

checksum=checksum+d

count=count+1

n=int(n/10)

if(checksum%10==0):

is_valid=1

return is_valid, checksum

def main():

test_id = "176248"

is_valid, checksum = validateID(test_id)

if(is_valid==1):

print(test_id+" is valid and checksum is "+str(checksum))

else:

print(test_id+" is not valid and checksum is "+str(checksum))

test_id = "79927398713"

is_valid, checksum = validateID(test_id)

if(is_valid==1):

print(test_id+" is valid and checksum is "+str(checksum))

else:

print(test_id+" is not valid and checksum is "+str(checksum))

test_id = "6080320539447211"

is_valid, checksum = validateID(test_id)

if(is_valid==1):

print(test_id+" is valid and checksum is "+str(checksum))

else:

print(test_id+" is not valid and checksum is "+str(checksum))

return

if __name__ == '__main__':

main()

7 0

3 years ago

Rob has 7 ones, nick has 5 ones.they put all their ones together.what number did they make?

Anni [7]

7+5= 12 ones. Or one ten and 2 ones. They made 12.

3 0

4 years ago

Read 2 more answers

225 is the percent 300

NISA [10]

I assume you are asking what percent is 255 out 0f 300. And the answer to that is 85%

6 0

4 years ago

14+16+49+10<br><br>Hint: Add what will get a tens number first!

Leviafan [203]

<span>89 is the correct answer.

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8 0

4 years ago

Read 2 more answers

A survey showed that 82% of adults need correction (eyeglasses, contacts, surgery, etc.) for their eyesight. If 15 adults are

Zigmanuir [339]

Answer:

$P(X \leq 1)= P(X=0) +P(X=1)$

And using the probability mass function we can find the individual probabiities

$P(X=0)=(15C0)(0.82)^0 (1-0.82)^{15-0}=6.75x10^{-12}$

$P(X=1)=(15C1)(0.82)^1 (1-0.82)^{15-1}=4.61x10^{-10}$

And replacing we got:

$P(X \leq 1)= P(X=0) +P(X=1)= 4.68x10^{-10}$

And for this case yes we can conclude that 1 a significantly low number of adults requiring eyesight correction in a sample of 15 since the probability obtained is very near to 0

Step-by-step explanation:

Let X the random variable of interest "number of adults who need correction", on this case we now that:

$X \sim Binom(n=15, p=0.82)$

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:

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

We want to find this probability:

$P(X \leq 1)= P(X=0) +P(X=1)$

And using the probability mass function we can find the individual probabiities

$P(X=0)=(15C0)(0.82)^0 (1-0.82)^{15-0}=6.75x10^{-12}$

$P(X=1)=(15C1)(0.82)^1 (1-0.82)^{15-1}=4.61x10^{-10}$

And replacing we got:

$P(X \leq 1)= P(X=0) +P(X=1)= 4.68x10^{-10}$

And for this case yes we can conclude that 1 a significantly low number of adults requiring eyesight correction in a sample of 15 since the probability obtained is very near to 0

6 0

3 years ago