Ask question

Ask question

All categories

3 years ago

14

In a coin collection,the number of dimes is 6 more than twice the number of nickels if the collection has 36 dimes how many nick

els are in the collection

2 answers:

gulaghasi [49]3 years ago

4 0

15 nickels, hope this helps :)

-BARSIC- [3]3 years ago

4 0

15 nickles! i hope thats how you spell nickles lol

You might be interested in

a boat is bringing a diver up at 25 feet per minute, the diver is currently at 200 feet, where was he 10 minutes ago ???

Sidana [21]

Answer:he was at -50 feet

Step-by-step explanation:

If he is going 25 feet per minute then you do 25x10 winch is 250 to see how far they have been total then since hes at 200 feet now you subtract 250 from 200 and you get your answer

5 0

3 years ago

Which equation in slope-intercept form represents a line that is parallel to y=−4x−5 and passes through the point (0,0)?

IrinaK [193]

Answer:

$\displaystyle y = -4x$

Step-by-step explanation:

0 = 0 ± b

$\displaystyle 0 = b \\ \\ y = -4x$

* Parallel Lines have SIMILAR <em>RATE OF CHANGES</em> [<em>SLOPES</em>], so −4 remains the way it is.

I am joyous to assist you anytime.

6 0

3 years ago

X - (7) = 13 help please

Bond [772]

Answer:

x=20

Step-by-step explanation:

inverse operation

add 7 to both sides

x=20

7 0

3 years ago

Read 2 more answers

PLS HELP PLS PLS!!!!!!

jenyasd209 [6]

Answer:

m > 1

Step-by-step explanation:

Minus 6 from both side

m = 1

4 0

3 years ago

The following results come from two independent random samples taken of two populations.

photoshop1234 [79]

Answer:

$(a)\ \bar x_1 - \bar x_2 = 2.0$

$(b)\ CI =(1.0542,2.9458)$

$(c)\ CI = (0.8730,2.1270)$

Step-by-step explanation:

Given

$n_1 = 60$ $n_2 = 35$

$\bar x_1 = 13.6$ $\bar x_2 = 11.6$

$\sigma_1 = 2.1$ $\sigma_2 = 3$

Solving (a): Point estimate of difference of mean

This is calculated as: $\bar x_1 - \bar x_2$

$\bar x_1 - \bar x_2 = 13.6 - 11.6$

$\bar x_1 - \bar x_2 = 2.0$

Solving (b): 90% confidence interval

We have:

$c = 90\%$

$c = 0.90$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.90$

$\alpha = 0.10$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.10/2}$

$z_{\alpha/2} = z_{0.05}$

The z score is:

$z_{\alpha/2} = z_{0.05} =1.645$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.645 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.645 * \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.645 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.645 * \sqrt{0.3306}$

$2.0 \± 0.9458$

Split

$(2.0 - 0.9458) \to (2.0 + 0.9458)$

$(1.0542) \to (2.9458)$

Hence, the 90% confidence interval is:

$CI =(1.0542,2.9458)$

Solving (c): 95% confidence interval

We have:

$c = 95\%$

$c = 0.95$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.95$

$\alpha = 0.05$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.05/2}$

$z_{\alpha/2} = z_{0.025}$

The z score is:

$z_{\alpha/2} = z_{0.025} =1.96$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.96 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.96* \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.96 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.96* \sqrt{0.3306}$

$2.0 \± 1.1270$

Split

$(2.0 - 1.1270) \to (2.0 + 1.1270)$

$(0.8730) \to (2.1270)$

Hence, the 95% confidence interval is:

$CI = (0.8730,2.1270)$

8 0

3 years ago