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3 years ago

A student saves $3 dollars on August 1, $5 on August 2, $7 on August 3, and so on. How much will she save in August 20 ?

2 answers:

5 0

Here is your answer

Money saved on each successive day is-

$3, &5, $7....

Clearly it forms an AP,

where

a1= 3

common difference, d= 2

n=20

So,

using formula

Tn= a1+(n-1)d

T20= 3+(20-1)2

= 3+ 19×2

= 3+ 38

=41

So, money saved on August 20= $41

Sum of money saved upto August 20 =
n/2 (a1+T20)
= 20/2 (3+41)
= 10× 44
= $440

HOPE IT IS USEFUL

Nookie1986 [14]3 years ago

4 0

$41. It increases by $2 each day. Formula 2n+1 :)

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I need help finding y

kolezko [41]

to solve for y, we must multiply by t on both side to get ride of the t in the denominator. By doing this, we will get:

${t}^{3}$

on the right side.

Subtracting 3, and we have successfully isolated y.

It would be impossible to get a quantitative value for y if we don't know the value of t.

4 0

3 years ago

Read 2 more answers

At the bottom is a completed addition problem, with all the digits replaced by letters. Every letter represents a single digit a

Rashid [163]

Answer:

T = 5 (H = 7, M = 1); T = 7 (H = 4, M = 2); T = 9 (H = 1, M = 3)

Step-by-step explanation:

We know MH+MH+MH=TM

MH is a two digit number, and TM is also a two digit number.

MH+MH+MH=3*(MH)<100

3M<=9, meaning that M could be 1, 2, or 3.

If M=1, MH+MH+MH=3(10+H)=30+3H=TM=10*T+1

30+3H=10T+1, so 3H must include 1 in one's digit.

The only possibility is that 3H = 21, then H = 7

MH+MH+MH=17+17+17=TM=51, T = 5

If M=2, MH+MH+MH=3(20+H)=60+3H=TM=10T+2

60+3H=10T+2, so 3H must include 2 in one's digit.

The only possibility is 3H = 12, then H = 4

MH+MH+MH=24+24+24=72, T = 7

If M=3, MH+MH+MH=3(30+H)=90+3H=TM=10T+3

90+3H=10T+3, so 3H must include 3 in one's digit.

The only possibility is 3H = 3, then H =1

MH+MH+MH=31+31+31=93, T=9

4 0

3 years ago

I got zero, but that’s wrong.

meriva

Answer:

Expected value is 4,000

Step-by-step explanation:

To find expected value ⇒ multiply the value by it's probability

40% × ( -25,000 ) = - 10,000

Breaking means neither add nor subtract a given amount

20% × 0 = 0

35% × 40,000 = 14,000

∴ Expected value = -10,000 + 14,000 = 4,000

LOSS: -10,000

BREAK EVEN: 0

WIN: 14,000

Expected value is 4,000

hope this helps....

3 0

3 years ago

What is the area of the shape given above?!? Someone please help this is due today

LenaWriter [7]

Answer:

15 m

Step-by-step explanation:

First split the shape into a square and a triangle, then find the area of the square by multipliyng the sides to get.

Next find the length of the traignel by subtracting 3 from 9 to get for then find the area of the triangle by multiplying 4 by 3 to get 12 then divide by 2 to get the area which is 6.

Finally add the 9 to the 6 to get 15

3 * 3 = 9

9 - 3 = 4

4 * 3 = 12/2 = 6

6+9=15

7 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

3 years ago