Which are similar to AJKL? Label each Yes or No. L м ALKM AJLM ΔKML

Pamela ran 7/10 mile on Tuesday and 3/8 mile on Thursday. How much fatter did she run on Tuesday ?

Len [333]

Answer:

$Difference = \frac{13}{40}$

Step-by-step explanation:

Given

$Thursday = \frac{3}{8}\ mile$

$Tuesday = \frac{7}{10}\ mile$

Required

How much farther is she on Tuesday

To do this, we simply calculate the difference between distance covered on both days.

$Difference = Tuesday-Thursday$

$Difference = \frac{7}{10} - \frac{3}{8}$

Take LCM

$Difference = \frac{28- 15}{40}$

$Difference = \frac{13}{40}$

Hence, she ran 13/40 mile farther on Tuesday

3 0

2 years ago

Peter is writing a number in scientific notation.The number is grater that ten million and less that one hundred million.Which e

OLga [1]

Answer:

$M = a \times 10^7$ where 1 < a ≤ 9

The EXPONENT USED BY PETER is 7 in all the possible cases.

Step-by-step explanation:

Let us assume the needed number is M.

Also, given:

The needed number is GREATER than 10 Million.

⇒ M > 10 Million

⇒ M > 10,000,000

⇒ M > $1 \times 10^7$ ...... (1)

The needed number is SMALLER than 100 Million.

⇒ M < 100 Million

⇒ M < 100,000,000

⇒ M < $1 \times 10^8$ ...... (2)

Comparing (1) and (2) we can see that M can have more than 7 zeroes after 1 but LESS than 8 zeroes after 1.

So, the possible values of M can be:

$M = a \times 10^7$ where 1 < a ≤ 9

So, the EXPONENT USED BY PETER is 7 in all the possible cases.

5 0

2 years ago

Answer...........................

Usimov [2.4K]

Answer:

9.9 years

Step-by-step explanation:

A = P e ^(rt)

Where A is the amount in the account

P is the amount invested

R is the interest rate

t is the time

P = 8500

r =7% = .07

A = 17000

Substituting into the equation

17000=8500 e^(.07t)

Divide each side by 8500

17000/8500=8500/8500 e^(.07t)

2 = e^(.07t)

Take the natural log of each side

ln (2) = ln e^(.07t)

ln(2) = .07t

Divide each side by .07

ln(2)/.07 = .07t/.07

ln(2)/.07 = t

9.902102579=t

Rounding to one decimal place

9.9 years

4 0

2 years ago

Read 2 more answers

$175 for 5 football game tickets how much is per ticket

vodomira [7]

For one ticket is $35

8 0

2 years ago

Read 2 more answers

A stationary store has decided to accept a large shipment of ball-point pens if an inspection of 19 randomly selected pens yield

Romashka [77]

Given Information:

Probability of shipment accepted = p = 5%

Probability of shipment not accepted = q = 95%

Total number of pens = n = 19

Required Information:

Probability of shipment being accepted with no more than 2 defective pens = P( x ≤ 2) = ?

Answer:

P( x ≤ 2) = 0.933

Step-by-step explanation:

The given problem can be solved using Bernoulli distribution which is given by

P(n, x) = nCx pˣqⁿ⁻ˣ

The probability of no more than 2 defective pens means

P( x ≤ 2) = Probability of 0 defective pen + Probability of 1 defective pen + Probability of 2 defective pens

P( x ≤ 2) = P(0) + P(1) + P(2)

For P(0) we have p = 0.05, q = 0.95, n = 19 and x = 0

P(0) = 19C0(0.05)⁰(0.95)¹⁹

P(0) = (1)(1)(0.377)

P(0) = 0.377

For P(1) we have p = 0.05, q = 0.95, n = 19 and x = 1

P(1) = 19C1(0.05)¹(0.95)¹⁸

P(1) = (19)(0.05)(0.397)

P(1) = 0.377

For P(2) we have p = 0.05, q = 0.95, n = 19 and x = 2

P(2) = 19C2(0.05)²(0.95)¹⁷

P(2) = (171)(0.0025)(0.418)

P(2) = 0.179

Therefore, the required probability is

P( x ≤ 2) = P(0) + P(1) + P(2)

P( x ≤ 2) = 0.377 + 0.377 + 0.179

P( x ≤ 2) = 0.933

P( x ≤ 2) = 93.3%

Therefore, the probability that this shipment is accepted with no more than 2 defective pens is 0.933.

8 0

2 years ago