All categories

3 years ago

Help me solve this problem!!

Mathematics

2 answers:

MariettaO [177]3 years ago

6 0

Answer: Approximately 6573481.358 Hispanic businesses

Step-by-step explanation:

In 2009, the number of Hispanic owned businesses was 1.15 million

In 2013, the number of Hispanic owned businesses was 1.52 million

Since the growth is exponential, we would apply

Tt = ar^t-1

Where Tt is the number of Hispanic owned businesses after t years

a is the number of Hispanic owned businesses in 2009

r is the common ratio or exponential growth rate of the number of Hispanic owned business

at 2013 , t = 5(t= 0 in 2009 to t= 5 in 2013)

1520000 = 1150000 × r^5-1

1520000 / 1150000 = r^4

152/115 = r^4

1.32174 = r^4

Taking 4th root of both sides,

r = 1.07222

We want to determine the number of Hispanic owned business that would be there in 2034. It becomes

T2034 = 1150000 × 1.07222^25

T2034 = 1150000 × 5.7160

T2034 = 6573481.3579901

Approximately 6573481.358 Hispanic businesses

grigory [225]3 years ago

3 0

Answer:

f(t) = 1.15(152/115)^(t/4)
f(25) ≈ 6.57 million

Step-by-step explanation:

(a) The ratio in 4 years is 1.52/1.15 = 152/115. So, the exponential function can be written using the form ...

f(t) = (initial value) × (ratio in period)^(t/(length of period))

Here, the initial value is 1.15 million, the ratio in the period of 4 years is 152/115, so the exponential function is ...

f(t) = 1.15(152/115)^(t/4)

Since no calculations were done, no rounding is necessary.

(b) After 25 years, this formula predicts the number of Hispanic owned businesses to be ...

f(25) = 1.15(152/115)^(25/4) ≈ 6.57 . . . . . million

_____

Alternate expressions of the exponential function

We can divide out the ratio we used and write the function using that:

f(t) = 1.15·1.321739^(t/4)

or we can take the 4th root of it to give ...

f(t) = 1.15·1.0722263^t

or we can take the log of it to get ...

f(t) = 1.15e^(0.069737t)

and we can even fold the initial constant into the exponent:

f(t) = e^(0.069737t +0.139762)

or rearrange to a slightly different form:

f(t) = e^(0.069737(t +2.004126))

You might be interested in

A certain forest covers an area of 2000 km. Suppose that each year this area decreases by 3.75%. What will the area be after Il

nasty-shy [4]

Exponential decay formula is Total = starting value x (1- percentage)^ time

Total = 2000(1- 0.0375)^11

Total = 2000(0.9625)^11

Total = 1313.524 km

Rounded to nearest km = 1314 km

5 0

4 years ago

a positive number is 7 times another number if 3 is added to both the numbers then one of the new number becomes 5 by 2 times th

Neko [114]

Answer:

7 and 1

Step-by-step explanation:

Let the numbers be a and b.

A positive number is 7 times another number:

a = 7b

If 3 is added to both the numbers then one of the new number becomes 5 by 2 times the other new number:

a+3 = 5/2 × (b +3)

To solve this we substitute a with 7b in the second equation:

7b + 3 = 5/2 × (b +3) ⇒ multiplying both sides by 2
14b + 6 = 5b + 15 ⇒ collecting like terms
14b - 5b = 15 - 6
9b = 9
b = 1 ⇒ solved for b

Then, finding a:

a= 7b
a=7*1
a= 7 ⇒ solved for a

So the numbers are 7 and 1

3 0

3 years ago

Does the histogram appear to depict data that have a normal distribution?

grin007 [14]

Answer:

B. The histogram appears to roughly approximate a normal distribution. The frequencies generally increase to a maximum and then decrease, and the histogram is symmetric.

Step-by-step explanation:

The Graph of Normal Distribution is like a bell-shaped. Here the value of y is less for the lower value of x and then the value of y is increased for a larger value of x, but after some time value of y is again getting decrease as the value of x increases. For the Histogram to appear to a normal distribution, the graph of histogram must have the same nature. Thus option B is only the correct option.

The histogram is made up of columns bar with no gaps between bars with different labels of numeric data of different heights shows the size of the group of different labels.

3 0

3 years ago

Read 2 more answers

Solve the problem below and show your work

ehidna [41]

The derivative of the function is 2x/(3+x²).

<h3>What is Derivative?</h3>

The derivative is the instantaneous rate of change of a function with respect to one of its variables.

Given function:

y= ln(3+x²)

Differentiating and applying chain rule

=1/(3+x²) d/dx(3+x²)

=1/(3+x²) (2x)

= 2x/(3+x²)

Hence, the derivative of the function is 2x/(3+x²).

Learn more about derivative here:

brainly.com/question/124529

#SPJ1

6 0

3 years ago

Land's Bend sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the intern

Naya [18.7K]

Answer:

80% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases ( $\mu_1-\mu_2$ ) is [-9.132 , 23.332].

Step-by-step explanation:

We are given that a random sample of 7 sales receipts for mail-order sales results in a mean sale amount of $81.70 with a standard deviation of $18.75.

A random sample of 11 sales receipts for internet sales results in a mean sale amount of $74.60 with a standard deviation of $28.25.

Firstly, the Pivotal quantity for 80% confidence interval for the difference between population means is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t__n__1-_n__2-2$

where, $\bar X_1$ = sample mean sales receipts for mail-order sales = $81.70

$\bar X_2$ = sample mean sales receipts for internet sales = $74.60

$s_1$ = sample standard deviation for mail-order sales = $18.75

$s_2$ = sample standard deviation for internet sales = $28.25

$n_1$ = size of sales receipts for mail-order sales = 7

$n_2$ = size of sales receipts for internet sales = 11

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(7-1)\times 18.75^{2} +(11-1)\times 28.25^{2} }{7+11-2} }$ = 25.11

Here for constructing 80% confidence interval we have used Two-sample t test statistics as we don't know about population standard deviations.

So, 80% confidence interval for the difference between population means, ( $\mu_1-\mu_2$ ) is ;

P(-1.337 < $t_1_6$ < 1.337) = 0.80 {As the critical value of t at 16 degree

of freedom are -1.337 & 1.337 with P = 10%}

P(-1.337 < $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 1.337) = 0.80

P( $-1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ${(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}$ < $1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.80

P( $(\bar X_1-\bar X_2)-1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ( $\mu_1-\mu_2$ ) < $(\bar X_1-\bar X_2)+1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.80

80% confidence interval for ( $\mu_1-\mu_2$ ) =

[ $(\bar X_1-\bar X_2)-1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ , $(\bar X_1-\bar X_2)+1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ]

= [ $(81.70-74.60)-1.337 \times {25.11 \times \sqrt{\frac{1}{7} +\frac{1}{11} } }$ , $(81.70-74.60)+1.337 \times {25.11 \times \sqrt{\frac{1}{7} +\frac{1}{11} } }$ ]

= [-9.132 , 23.332]

Therefore, 80% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases ( $\mu_1-\mu_2$ ) is [-9.132 , 23.332].

4 0

3 years ago