Could I get a hand with this, please?

The function below describes the number of students who enrolled at a university, where f(t) represents the number of students a

dem82 [27]

Answer:

18,500 , 1 , 1.03

Step-by-step explanation:

I think your question is missed of key information, allow me to add in and hope it will fit the original one.

f(t) =18,500(1.03)^t

1st blank:

19,055

1.03

3

18,500

2nd blank:

t years

2 years

3 years

1 year

3rd blank:

19,055

3

18,500

1.03

My answer:

Given that: f(t) =18,500* $1.03^{t}$

1st blank:

Initial value when t = 0, so we have:

f(0) =18,500* $1.03^{0}$

$f(0) =18,500(1)\\ f(0) =18,500$

So we choose D for 1st blank

2nd and 3rd blank:

Because it's an exponential function f(t) =18,500* $1.03^{t}$ , with every value of t increment f(t) increase by a factor of 1.03 . So Every 1 year the number of students who enroll at the university increases by a factor of 1.03

=> 2nd blank: 1

=> 3rd blank: 1.03

Hope it will find you well.

6 0

3 years ago

Expand 4(-2y + 3). 2 y + 7 -8 y + 12 2 y + 12 -8 y + 3

mrs_skeptik [129]

Answer: -8y+12

Step-by-step explanation: Multiply using the distributive property.

Hope this helps.

8 0

3 years ago

Find the equation of the axis of symmetry of the graph of y = (x + 3)^2 - 5

Temka [501]

Equation of the parabola: y = ax^2 + bx + c. Find a, b, and c.
x of axis of symmetry:
x
=
−
b
2
a
=
3
-> b = -6a
Writing that the graph passing at point (1, 0) and point (4, -3):
(1) 0 = a + b + c -> c = - a - b = - a + 6a = 5a
(2) -3 = 16a + 4b + c --> -3 = 16a - 24a + 5a = -3a --> a = 1
b = -6a = -6; and c = 5a = 5
y
=
x
2
−
6
x
+
5

Check with x = 1: -> y = 1 - 6 + 5 = 0. OK

3 0

3 years ago

Can someone pls explain to me how to do this (^｡^)

34kurt

Answer:

Area: $\frac{8}{15}$ feet²

Perimeter: $\frac{44}{15} =2 \frac{14}{15}$ feet

Step-by-step explanation:

The formula of finding the area of a rectangle or square is width multiplied by length.

The formula of find the perimeter of a rectangle or square is width+width+length+length

Area: $\frac{4}{5} *\frac{2}{3} = \frac{8}{15}$ feet²

Perimeter:

$\frac{4}{5} +\frac{4}{5} +\frac{2}{3} +\frac{2}{3} =$

$\\\\\frac{12}{15} +\frac{12}{15} +\frac{10}{15} +\frac{10}{15} =\\\\\frac{24}{15} +\frac{20}{15} =\\$

$\frac{44}{15} =2 \frac{14}{15}$

4 0

3 years ago

Vitamin D, whether ingested as a dietary supplement or produced naturally when sunlight falls on the skin, is essential for stro

tigry1 [53]

Answer:

a) The 98% confidence interval would be given (0.182;0.218).

b) We are 98% confident that the true proportion of of England people who are deficient in Vitamin D is between 0.182 and 0.218.

c) If repeated samples were taken and the 98% confidence interval computed for each sample, 98% of the intervals would contain the population proportion.

d) Yes since the confidence interval not contains the value 0.25, we can refute the claim at 2% of significance.

Step-by-step explanation:

Data given and notation

n=2700 represent the random sample taken

X represent the people in England who are deficient in Vitamin D

$\hat p=0.2$ estimated proportion of England people who are deficient in Vitamin D

$\alpha=0.02$ represent the significance level

Confidence =0.98 or 98%

z would represent the statistic (variable of interest)

p= population proportion of England people who are deficient in Vitamin D

Part a

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 98% confidence interval the value of $\alpha=1-0.98=0.02$ and $\alpha/2=0.01$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.33$

And replacing into the confidence interval formula we got:

$0.20 - 2.33 \sqrt{\frac{0.2(1-0.2)}{2700}}=0.182$

$0.20 + 2.33 \sqrt{\frac{0.2(1-0.2)}{2700}}=0.218$

And the 98% confidence interval would be given (0.182;0.218).

Part b

We are 98% confident that the true proportion of of England people who are deficient in Vitamin D is between 0.182 and 0.218.

Part c

If repeated samples were taken and the 98% confidence interval computed for each sample, 98% of the intervals would contain the population proportion.

Part d

Yes since the confidence interval not contains the value 0.25, we can refute the claim at 2% of significance.

3 0

3 years ago