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2 years ago

What is 4/ 15 divided by 10/ 13

Mathematics

2 answers:

zloy xaker [14]2 years ago

8 0

Hello,

The answer is 26/75 or 0.34666666667 if you make all of the fractions decimals then divide you can fine your answer

hope this helps

plz mark me as brainliest

Alja [10]2 years ago

5 0

Answer:

26/75

Step-by-step explanation:

4 / 15 / 10 / 13

= 4/15 * 13/10

= 52 / 150

= 26/75 answer

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Deep Blue, a deep sea fishing company, bought a boat for $250,000. After 9 years, Deep Blue plans to sell it for a scrap value o

zhuklara [117]

Answer:

Therefore, we use the linear depreciation and we get is 17222.22 .

Step-by-step explanation:

From Exercise we have that is boat $250,000.

The straight line depreciation for a boat would be calculated as follows:

Cost boat is $250,000.

For $95,000 Deep Blue plans to sell it after 9 years.

We use the formula and we calculate :

(250000-95000)/9=155000/9=17222.22

Therefore, we use the linear depreciation and we get is 17222.22 .

8 0

2 years ago

What is the range of the function f(x) = 3x2 + 6x – 8?

yawa3891 [41]

F(x) is a quadratic equation with the x-side squared and a is positive which means that the graph of the function is a parabola facing up. The range of f(x) is given by {y|y ≥ k}, where k is the y-coordinate of the vertex.
$f(x)=3 x^{2} +6x-8$ , written in vertex form is
$y=3 (x+1)^{2} -11$ , where (h, k) = (-1, -11)
Therefore, range ={y|y ≥ -11}

5 0

2 years ago

Read 2 more answers

A homemade lip balm consists of coconut oil and beeswax. Coconut oil costs $0.50 per ounce and beeswax costs $2.00 per ounce. If

Alika [10]

Let

a--------> the total cost in dollars of $6$ ounces of coconut oil

b--------> the total cost in dollars of $5$ ounces of beeswax

we know that

$a=0.50 \frac{\$}{ounce}*6=\$3$

$b=2.00 \frac{\$}{ounce}*5=\$10$

therefore

the answer is

the total cost in dollars of $6$ ounces of coconut oil is $a=\$3$

the total cost in dollars of $5$ ounces ofbeeswax is $b=\$10$

7 0

2 years ago

Read 2 more answers

Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in

Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of Americans who decide to not go to college = 48%

n = sample of American adults = 331

p = population proportion of Americans who decide to not go to

college because they cannot afford it

Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.

So, 90% confidence interval for the population proportion, p is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

90% confidence interval for p = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ , $0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ ]

= [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$

$0.015 = 1.645 \times \sqrt{\frac{0.48(1-0.48)}{n} }$

$\sqrt{n} = \frac{1.645 \times \sqrt{0.48 \times 0.52} }{0.015}$

$\sqrt{n}$ = 54.79

n = $54.79^{2}$

n = 3001.88 ≈ 3002

Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

5 0

2 years ago

$5000 was invested into an investment that pays 4.25% simple interest. The total value of the investment amounted to $6593.75. H

Basile [38]

Answer:

7.5 years

Step-by-step explanation:

P = $5000,

R = 4.25%,

A = $6593.75,

N =?

SI = A - P = 6593.75 - 5000 =$1593. 75

$\because \: SI = \frac{ PNR}{100} \\ \\ \therefore \: 1593.75 = \frac{5000 \times N \times 4.25}{100} \\ \\ \therefore \: 1593.75 \times 100 = 21,250 \times N \\ \\ N = \frac{159375}{21250} \\ \\ N = 7.5 \: years \:$

6 0

2 years ago