Ask question

Ask question

All categories

Ostrovityanka [42]

3 years ago

14

A total of $150000 is invested in two funds paying 6.25% and 6% simple interest. If the total interest for the year is $9212.50,

how much is invested at each rate?

1 answer:

Georgia [21]3 years ago

3 0

Answer:

$85,000 at 6.25%
$65,000 at 6.00%

Step-by-step explanation:

Let x represent the amount invested at 6.25%. Then the total interest earned is ...

0.0625x + 0.0600(150,000 -x) = 9212.50

0.0025x = 212.50 . . . . . . subtract 9000, collect terms

x = 212.50/0.0025 = 85,000

150,000 - 85,000 = 65,000 . . . . amount invested at the lower rate

$85,000 is invested at 6.25%; $65,000 is invested at 6%.

You might be interested in

How many hundreds are in this number 128

Nataliya [291]

1 hundreds
2 tens
8 ones

6 0

3 years ago

Help me solve this problem 98+p=56

Arlecino [84]

p=-42

move 98 to the right side of the equal sign which becomes -98

p=56-98

do the math

p=-42

3 0

3 years ago

Read 2 more answers

If there is 228 students in 12 classes, how many students are in 1 class?

stealth61 [152]

You would divide 228/12 = 19 per class.

3 0

2 years ago

A forest covers 43,000 acres. A survey finds that 0.7% of the forest is old-growth trees. How many acres of old-growth trees

Goryan [66]

Answer:

301 acres

Step-by-step explanation:

43,000 acres of forest

0.7% is old-growth

0.7% = 0.007

43,000 * 0.007 = 301 acres

7 0

3 years ago

Location is known to affect the number, of a particular item, sold by an automobile dealer. Two different locations, A and B, ar

yKpoI14uk [10]

Answer:

We conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

Step-by-step explanation:

We are given that Location A was observed for 18 days and location B was observed for 13 days.

On average, location A sold 39 of these items with a sample standard deviation of 8 and location B sold 49 of these items with a sample standard deviation of 4.

<em>Let </em> $\mu_1$ <em> = true mean number of sales at location A.</em>

<em /> $\mu_2$ = <em>true mean number of sales at location B</em>

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2\geq0$ or $\mu_1 \geq \mu_2$ {means that the true mean number of sales at location A is greater than or equal to the true mean number of sales at location B}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2$ or $\mu_1< \mu_2$ {means that the true mean number of sales at location A is fewer than the true mean number of sales at location B}

The test statistics that would be used here <u>Two-sample t test statistics</u> as we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n__1_-_n__2-2$

where, $\bar X_1$ = sample average of items sold at location A = 39

$\bar X_2$ = sample average of items sold at location B = 49

$s_1$ = sample standard deviation of items sold at location A = 8

$s_2$ = sample standard deviation of items sold at location B = 4

$n_1$ = sample of days location A was observed = 18

$n_2$ = sample of days location B was observed = 13

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(18-1)\times 8^{2}+(13-1)\times 4^{2} }{18+13-2} }$ = 6.64

So, <u><em>test statistics</em></u> = $\frac{(39-49)-(0)}{6.64 \times \sqrt{\frac{1}{18}+\frac{1}{13} } }$ ~ $t_2_9$

= -4.14

The value of t test statistics is -4.14.

Now, at 0.01 significance level the t table gives critical value of -2.462 at 29 degree of freedom for left-tailed test.

<em>Since our test statistics is less than the critical values of t as -2.462 > -4.14, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><u><em>we reject our null hypothesis</em></u><em>.</em>

Therefore, we conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

3 0

3 years ago