What is equivalent to 1+1 because i do not know

A supermarket finds that the number of boxes of a new cereal sold increases each week. in the first week, only 16 boxes of the c

vaieri [72.5K]

For this case, the arithmetic sequence is given by:
an = 17n-1
Let's check it:
First week:
n = 1:
a1 = 17 (1) -1
a1 = 17-1
a1 = 16
Second week:
n = 2:
a2 = 17 (2) -1
a2 = 34-1
a2 = 33
Third week:
n = 3:
a3 = 17 (3) -1
a3 = 51-1
a3 = 50
Answer:
the explicit rule for the arithmetic sequence that defines the number of boxes of cereal sold in week n is:
an = 17n-1

6 0

3 years ago

Double the product of 4y and 3b if y=2 and b=1.

Aleksandr-060686 [28]

Answer:

48

Step-by-step explanation:

Double the product of 4y and 3b

2 * ((4y) * (3b))

y = 2 and b = 1

2 * ((4 * 2) * (3 * 1))

2 * (8 * 3)

2 * 24

48

3 0

2 years ago

3. Find x, y, u, and v satisfying the system of four equations: x + 7y + 3v + 5u = 16 8x + 4y + 6v + 2u = -16 2x + 6y + 4v + 8u

Sophie [7]

Answer:356

Step-by-step explanation:

8 0

3 years ago

A juice pitcher holds 1.5 gallons of liquid. How many 8-ounce glasses of juice can be poured from a full pitcher? (1 gallon = 12

larisa86 [58]

$1gallon=128ounces\\To\ receive\ volume\ of\ pitcher\ in\ onces\ we\ need\ to\ multiply\ its\\volume\ in\ gallons\ by\ ounces\ in\ 1\ gallon.\\1,5g=1,5*128=192o\\Now\ to\ receive\ answer\ we\ need\ to\ divide\ volume\ of\ pitcher\ in\\ounces\ by\ volume\ of\ glass.\\192:8=24\\\\We\ need\ to\ 24\ glasses.$

6 0

3 years ago

Suppose that an airline overbooks seats on their flights. In particular, it sells 300 tickets for a flight when there are only 2

vladimir1956 [14]

Using the normal approximation to the binomial, it is found that there is a 0.994 = 99.4% probability that we will have enough seats for everyone who shows up.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.
After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$ .

In this problem:

15% do not show up, so 100 - 15 = 85% show up, which means that $p = 0.85$ .
300 tickets are sold, hence $n = 300$ .

The mean and the standard deviation are given by:

$\mu = np = 300(0.85) = 255$

$\sigma = \sqrt{np(1-p)} = \sqrt{300(0.85)(0.15)} = 6.185$

The probability that we will have enough seats for everyone who shows up is the probability of at most 270 people showing up, which, using continuity correction, is $P(X \leq 270 + 0.5) = P(X \leq 270.5)$ , which is the p-value of Z when X = 270.5.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{270.5 - 255}{6.185}$

$Z = 2.51$

$Z = 2.51$ has a p-value of 0.994.

0.994 = 99.4% probability that we will have enough seats for everyone who shows up.

A similar problem is given at brainly.com/question/24261244

8 0

2 years ago