A software company writes 1/10 of their code in 1/2 a month. How much can they do in one month

If Rita receives $45.34 interest for a deposit earning 4% simple interest for 160 days, what is the amount of her deposit? (Roun

vovangra [49]

Answer:

6.4

Step-by-step explanation:

5 0

3 years ago

Please help

Aloiza [94]

A=10x(15x)-p(4x)^2

A=150x^2-16px^2

A=(150-16p)x^2

6 0

3 years ago

There is a boardwalk game at Point Pleasant where you are blindfolded to throw darts at a board full of balloons. Each time a da

arlik [135]

The given number of balloons, green = 10, purple = 4, red = 5, tie-dye = 2, black = 3, gives;

a. 5/138

b. 25/276

c. 5/1012

d. 35/46

<h3>How can the different probabilities be calculated mathematically?</h3>

Given parameters;

Number of balls;

Green = 10

Purple = 4

Red = 5

Tie-dye = 2

Black = 3

Mode of selection = Without replacement

Number of balloons = 10+4+5+2+3 = 24

a. Probability of popping a red balloon = 5/24

Probability of popping a second red balloon = 4/23

Therefore;

Probability of popping two reds consecutively = 5/24 × 4/23 = 5/138

b. Probability of popping a red balloon = 5/24

Probability of popping a green balloon next = 10/23

Therefore;

Probability of popping a red and then a green balloon = 5/24 × 10/23 = 25/276

c. Probability that the first balloon that pops is a red = 5/24

Next balloon is a black = 3/23

Third balloon is red = 4/22

The probability, <em>P</em>, is therefore;

P = 5/24 × 3/23 × 4/22 = 5/1012

d. The probability that the first balloon is a tie-dye = 2/24 = 1/12

Therefore;

Probability that the first balloon is not a tie-dye = 1 - 1/12 = 11/12

Probability that the second balloon is not a tie-dye = 21/23

Similarly;

Probability that the third balloon is not a tie-dye = 20/22 = 10/11

Which gives;

The probability, <em>P</em>, of popping anything but a tie-dye on three consecutive throws is therefore;

P = 11/12 × 21/23 × 10/11 = 35/46

Learn more about probability theory in mathematics?

brainly.com/question/13604758

#SPJ1

7 0

1 year ago

he amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and s

sp2606 [1]

Complete question:

He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is

a) less than 8 minutes

b) between 8 and 9 minutes

c) less than 7.5 minutes

Answer:

a) 0.0708

b) 0.9291

c) 0.0000

Step-by-step explanation:

Given:

n = 47

u = 8.3 mins

s.d = 1.4 mins

a) Less than 8 minutes:

$P(X$

P(X' < 8) = P(Z< - 1.47)

Using the normal distribution table:

NORMSDIST(-1.47)

= 0.0708

b) between 8 and 9 minutes:

P(8< X' <9) = $[\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}]$

= P(-1.47 <Z< 6.366)

= P( Z< 6.366) - P(Z< -1.47)

Using normal distribution table,

$NORMSDIST(6.366)-NORMSDIST(-1.47)$

0.9999 - 0.0708

= 0.9291

c) Less than 7.5 minutes:

P(X'<7.5) = $P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}]$

P(X' < 7.5) = P(Z< -3.92)

NORMSDIST (-3.92)

= 0.0000

3 0

3 years ago

I try to solve it but I didn't get any of the choices <br> Thank you for helping me

Natasha2012 [34]

The answer is 7. You can look at it like this. The two triangles are similar and the larger one's sides are twice the size of the smaller one's. So then we know GF is twice the size of JK.
2(2y+5)=5y+3
4y+10=5y+3
y=7

8 0

3 years ago