All categories

4 years ago

Noelle has 5/6 of a yard of purple ribbon and 9/10 of a yard of Pink ribbon. How mucho ribbon does se have altogether?

1 answer:

8 0

Well you need to find a common denominator which 30 is the closest one. 5/6 becomes 25/30, 9/10 becomes 27/30. 27/30+ 25/30 = 52/30. To simplify that you divide by 2 to get 26/15. It is no longer divisible to a whole number after that point.

You might be interested in

Suppose you can afford only $200 a month in car payments and your best loan option is a 60-month loan at 3%. How much money coul

g100num [7]

Answer:

$11,130.47

Step-by-step explanation:

The amortization formula can be used. It tells you the monthly payment amount A for some principal P, interest rate r, and n payments.

A = P(r/12)/(1 -(1 +r/12)^(-n))

Filling in your values, we get ...

200 = P(.03/12)/(1 -(1 +.03/12)^-60) = P(.0025)/(1 -1.0025^-60)

P = 200(1 -1.0025^-60)/.0025 ≈ 200×55.6523577

P ≈ 11,130.47

The present value of the loan is $11,130.47.

8 0

3 years ago

Construct a 99% confidence interval for the population mean, mu. Assume the population has a normal distribution. A group of 1

Zarrin [17]

Answer:

99% confidence interval for the population mean is [19.891 , 24.909].

Step-by-step explanation:

We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

Assuming the population has a normal distribution.

Firstly, the pivotal quantity for 99% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean age of selected students = 22.4 years

s = sample standard deviation = 3.8 years

n = sample of students = 19

$\mu$ = population mean

<em>Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.</em>

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.878 < $t_1_8$ < 2.878) = 0.99 {As the critical value of t at 18 degree of

freedom are -2.878 & 2.878 with P = 0.5%}

P(-2.878 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.878) = 0.99

P( $-2.878 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.878 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ) = 0.99

<u>99% confidence interval for</u> $\mu$ = [ $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ , $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ]

= [ $22.4 -2.878 \times {\frac{3.8}{\sqrt{19} }$ , $22.4 +2.878 \times {\frac{3.8}{\sqrt{19} }$ ]

= [19.891 , 24.909]

Therefore, 99% confidence interval for the population mean is [19.891 , 24.909].

6 0

3 years ago

What could be the value for x? 3/14 < x < 3/13

Ksju [112]

Answer:

14^3/100 i think im not 100% sure so if its wrong dont get mad

Step-by-step explanation:

8 0

3 years ago

Plzzzzzz help am stuck

juin [17]

50.17
..................

4 0

3 years ago

Read 2 more answers

David went to a fast-food shop for a meal in a late evening. Only steak, pork chop and grilled chicken were available for main c

Doss [256]

six

Step-by-step explanation:

the reason is this: let steak be (s), pork be (p), grilled chicken be (gc), tea be (t), and coffee be (c)

now, we want to find the total number of outcomes, and this cannot be determined by a simple roll of the dice. you have to count the sample space required for EVERY outcome, and it has to lie within the conditions. so, assuming David only takes 1 main course and 1 drink, he would be down to six option, which are :

1. s & t

2. s & c

3. p & t

4. p & c

5. gc & t

6. gc & c

5 0

3 years ago