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3 years ago

Which is the smaller number: -5 or -15?

Mathematics

1 answer:

m_a_m_a [10]3 years ago

7 0

Answer:

-15

Step-by-step explanation:

-15 is smaller because it is a negative number. -5 is closer to 0. -15 is farther away from 0.

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The total price of 5 pounds of bananas was $1.95. What was the price per pound?

Ira Lisetskai [31]

Answer:

Well you divide the total of 1.95/5

.39 cent

8 0

2 years ago

Read 2 more answers

For each function, what is the output of the given input? for f(x) = 4x 7, find f(4)

shusha [124]

I hope this helps you

x =4

f (4)= 4.4+7 = 16+7=23

f (4)= 4.4-7=16-7=9

6 0

2 years ago

1. take away five from twelve times f. 2. one-half of the sum of k and six 3.x squared minus the sum of 5 4.the sum of the produ

77julia77 [94]

The expressions formed are,

12f -5
(k+6)/2
x²-x+5
ab+3c
24xy+g

Formation of expressions in 1, 2 and 3:

In 1, twelve times f is, 12f

Taking away 5, it becomes (12f-5)

In 2, sum of k and 6 is, (k+6)

One-half of the above quantity is, (k+6)/2

In 3, sum of 5 with x is, (x+5)

Now, x squared minus the above expression indicates (x²-x+5)

Formation of expressions in 4 and 5:

In 4, product of a and b, is ab and 3 times c is 3c

Sum of the expressions evaluated in the previous statement = ab+3c

In 5, 24 times the product of x and y is, 24xy

Adding, g in the above computed expression, we get, 24xy+g

Learn more about expressions here:

brainly.com/question/17167810

#SPJ4

3 0

1 year ago

The expected number of typographical errors on a page of a certain magazine is .2. What is the probability that an article of 10

Pavel [41]

Answer:

a) The probability that an article of 10 pages contains 0 typographical errors is 0.8187.

b) The probability that an article of 10 pages contains 2 or more typographical errors is 0.0175.

Step-by-step explanation:

Given : The expected number of typographical errors on a page of a certain magazine is 0.2.

To find : What is the probability that an article of 10 pages contains

(a) 0 and (b) 2 or more typographical errors?

Solution :

Applying Poisson distribution,

$N\sim Pois(0.2)$

$P(N=r)=\frac{e^{-np}(np)^r}{r!}$

where, n is the number of words in a page

and p is the probability of every word with typographical errors.

Here, n=10 and E(N)=np=0.2

a) The probability that an article of 10 pages contains 0 typographical errors.

Substitute r=0 in formula,

$P(N=0)=\frac{e^{-0.2}(0.2)^0}{0!}$

$P(N=0)=\frac{e^{-0.2}}{1}$

$P(N=0)=e^{-0.2}$

$P(N=0)=0.8187$

The probability that an article of 10 pages contains 0 typographical errors is 0.8187.

b) The probability that an article of 10 pages contains 2 or more typographical errors.

Substitute $r\geq 2$ in formula,

$P(N\geq 2)=1-P(N$

$P(N\geq 2)=1-[P(N=0)+P(N=1)]$

$P(N\geq 2)=1-[\frac{e^{-0.2}(0.2)^0}{0!}+\frac{e^{-0.2}(0.2)^1}{1!}]$

$P(N\geq 2)=1-[e^{-0.2}+e^{-0.2}(0.2)]$

$P(N\geq 2)=1-[0.8187+0.1637]$

$P(N\geq 2)=1-0.9825$

$P(N\geq 2)=0.0175$

The probability that an article of 10 pages contains 2 or more typographical errors is 0.0175.

6 0

3 years ago

Before every flight, the pilot must verify that the total weight of the load is less than the maximum allowable load for the ai

ss7ja [257]

Answer:

The probability that the plane is oveloaded is P=0.9983.

The pilot should take out the baggage and send it in another plain or have less passengers in the plain to not overload.

Step-by-step explanation:

The aircraft will be overloaded if the mean weight of the passengers is greater than 163 lb.

If the plane is full, we have 41 men in the plane. This is our sample size.

The weights of men are normally distributed with a mean of 180.5 lb and a standard deviation of 38.2.

So the mean of the sample is 180.5 lb (equal to the population mean).

The standard deviation is:

$\sigma=\frac{\sigma}{\sqrt{N}} =\frac{38.2}{\sqrt{41}}=\frac{38.2}{6.4} =5.97$

Then, we can calculate the z value for x=163 lb.

$z=\frac{x-\mu}{\sigma}=\frac{163-180.5}{5.97}=\frac{-17.5}{5.97}= -2.93$

The probability that the mean weight of the men in the airplane is below 163 lb is P=0.0017

$P(\bar X$

Then the probability that the plane is oveloaded is P=0.9983:

$P(overloaded)=1-P(X$

The pilot should take out the baggage or have less passengers in the plain to not overload.

7 0

3 years ago