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4 years ago

Round 13,365,328 to the nearest ten

Mathematics

1 answer:

Darina [25.2K]4 years ago

7 0

13,365,330 would be the answer

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Did Meg make it home on time if she spent 22 minutes at the library. 38 minutes at her grandmas house. She walked for 17 minutes

Tomtit [17]

Not enough info to answer

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3 years ago

Prove that; cosec²∅-2 = COS2∅ Sin2∅

MakcuM [25]

Answer:

here you go..............

7 0

3 years ago

Read 2 more answers

I will give brainiest to whoever answers correctly !!

NemiM [27]

Step-by-step explanation:

A) In order to calculate the interest , we can use the following formula :

$P=P_{0}.(1+i)^{t}$

Where P is the interest , P0 is the initial value, i is the interest and t is the time.

So , using P0 =600,i= 0.0005768 and t=27 ,we have :

$P=600.(1+0.0005768)^{27} \\P=600.(1.01569)\\P=609.41$

Calculating only the interest, we have :

$I=609.41-600\\I=9.41\\$

So the interest is $9.41.

B) He paid the final value calculated above, that is , $609.41.

PLEASE MARK ME AS BRAINLIEST

5 0

3 years ago

Steven wants to buy a $580 bicycle. Steven has no money saved, but will be able to deposit $30 into a savings account when he re

Bingel [31]

$30 + $30 = $60

$60 + $30 = $90

To pay back his sister the $80 he will to deposit money into his account for 3 weeks.

Since Steven would have $10 left over in his account from paying his sister all he needs now is $570.

You need to divide:

$570/$30 = 19

So Steven will need to deposit $30 into his bank account for 19 weeks to be able to buy the $580 bicycle.

Hope this Helps!!

8 0

4 years ago

Surface-finish defects in a small electric appliance occur at random with a mean rate of 0.3 defects per unit. find the probabil

mario62 [17]

Answer: Probability that a randomly selected unit will contain at least two surface- finish defect is 0.04.

Step-by-step explanation:

Since we have given that

Mean rate defects per unit = 0.3

Since we will use "Poisson distribution":

$P(X=K)=\frac{e^{-\lambda}\lambda^k}{k!}$

But we need to find the probability that a randomly selected unit will contain at least two surface-finish defect.

$P(X\geq 2)=1+P(X=0)+P(X=1)$

So,

$P(X=0)=\frac{e^{-0.3}0.3^0}{0!}=e^{-0.3}=0.74\\\\P(X=1)=\frac{e^{-0.3}\times 0.3}{1}=0.22$

so, it becomes,

$P(X>2)=1-(0.74+0.22)=1-0.96=0.04$

Hence, probability that a randomly selected unit will contain at least two surface- finish defect is 0.04.

4 0

4 years ago