All categories

2 years ago

10 divided 2-3(5times0)

Mathematics

1 answer:

artcher [175]2 years ago

3 0

Answer:

Step-by-step explanation:

10÷2-3 (5×0)

10÷2-3×0
5-0
5

You might be interested in

I need help with this.

Sveta_85 [38]

Answer:

check online for more information

6 0

1 year ago

Ok what next helppp i need 2

pickupchik [31]

Answer:

3-2

Step-by-step explanation

8 0

2 years ago

Which choice shows the students listed in order from who has the least amount of money to who has the greatest amount of money

exis [7]

Step-by-step explanation:

Should be Emily, Morgan, Kali, Rosa

6 0

2 years ago

(a) The Feelgood Gym has a monthly membership fee of $100 and it charges an addi-

Ierofanga [76]

If I use the gym for x hours in a month, the total cost in terms of x will be 1000 + x.

a. Monthly membership fee = $100

Additional amount charged = $1 per hour.

If I use the gym for x hours in a month, the total cost in terms of x will be:

= 1000 + 1(x)

= 1000 + x

b. Also, if the amount charged is $35 per month and $3.50 per hour, the total cost will be:

= 35 + 3.50(x)

= 35 + 3.5x

c. The number of hours that will give same cost will be:

1000 + x = 35 + 3.5x

Collect like terms.

1000 - 35 = 3.5x - x

965 = 2.5x

x = 965/2.5

x = 386

In 386 hours, the total cost will be the same.

Read related link on:

brainly.com/question/24515212

5 0

2 years ago

An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer. In a sample of 2000 individ

Tema [17]

Answer:

a) P=0.558

b) P=0.021

Step-by-step explanation:

We can model this random variable as a Poisson distribution with parameter λ=1/500*2000=4.

The approximate distribution of the number who carry this gene in a sample of 2000 individuals is:

$P(x=k)=\frac{\lambda^ke^{-\lambda}}{k!} =\frac{4^ke^{-4}}{k!}$

a) We can calculate that the approximate probability that between 4 and 9 (inclusive) as:

$P(4\leq x\leq 9)=\sum_{k=4}^9P(k)\\\\\\ P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\P(9)=4^{9} \cdot e^{-4}/9!=262144*0.0183/362880=0.013\\\\\\$

$P(4\leq x\leq 9)=\sum_{k=4}^9P(k)=0.195+0.156+0.104+0.060+0.030+0.013=0.558$

b) The approximate probability that at least 9 carry the gene is:

$P(x\geq9)=1-P(x\leq 8)\\\\\\$

$P(0)=4^{0} \cdot e^{-4}/0!=1*0.0183/1=0.018\\\\P(1)=4^{1} \cdot e^{-4}/1!=4*0.0183/1=0.073\\\\P(2)=4^{2} \cdot e^{-4}/2!=16*0.0183/2=0.147\\\\P(3)=4^{3} \cdot e^{-4}/3!=64*0.0183/6=0.195\\\\P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\$

$P(x\geq9)=1-P(x\leq 8)\\\\P(x\geq9)=1-(0.018+0.073+0.147+0.195+0.195+0.156+0.104+0.060+0.030)\\\\P(x\geq9)=1-0.979=0.021$

8 0

2 years ago