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1 year ago

help solve for initial value, the 1 unit growth/decay factor, the 1 unit percent change and the function for both parts a and b

Mathematics

1 answer:

Vladimir [108]1 year ago

5 0

we have that

i. Initial value

value of the function for t=0

<h2>the initial value is 23</h2>

ii.For t=1

k(t)=23(b)^t

k(1)=23(b)

29.67=23b

Part B

initial value ------>

iii. percent change

b=1+r

1.29=1+r

r=0.29

iv. the equation is

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4 years ago

Read 2 more answers

2. (04.01) Which point could be removed in order to make the relation a function? (4 points) {(-9, -8), (-8, 4), (0, -2), (4, 8)

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Answer:

We are given order pairs (–9, –8), (–{(8, 4), (0, –2), (4, 8), (0, 8), (1, 2)}.

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In the given order pairs, we can see that (0, –2) and (0, 8) order pairs has same x-coordinate 0.

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3 years ago

Quadrilateral ABCD is inscribed in circle 0 What is m<A

valentinak56 [21]

We know that
A quadrilateral is inscribed in a circle if and only if the opposite angles are supplementary. ( Inscribed Quadrilateral Theorem)
so
m∠B+m∠D=180°
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3 years ago

A public health official is planning for the supplyof influenza vaccine needed for the upcoming flu season. She wants to estimat

Marizza181 [45]

Answer:

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And rounded up we have that n=421

Step-by-step explanation:

We know that the sample proportion have the following distribution:

$\hat p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.1$ and $\alpha/2 =0.05$ . And the critical value would be given by:

$z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64$

The margin of error for the proportion interval is given by this formula:

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

And on this case we have that $ME =\pm 0.04$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

We assume that a prior estimation for p would be $\hat p =0.5$ since we don't have any other info provided. And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.64})^2}=420.25$

And rounded up we have that n=421

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3 years ago