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3 years ago

12

last week 267 people watched a football game. this week 574 people watched a football game how many more people watched this wee

ks game than last weeks

1 answer:

mel-nik [20]3 years ago

5 0

This is subtraction, what you do is subtract 267 from 574 (267 - 574 =___)

And your answer comes out to be 307.

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PLZ HELP ASAP!!

Triss [41]

<span>prove: the segment joining the midpoints of the two sides of a triangle is parallel to the third side. the coordinates are A(0,0)B(a,0)on the the x axis, and C(c,d), M</span>

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3 years ago

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If the cost of a graphing calculator is now three-fourths of what the calculator cost 5 years ago and the cost of the graphing c

Andru [333]

Answer:

58 1/2

Step-by-step explanation:

multiply by 3/4

3 0

2 years ago

Uninhibited growth can be modeled by exponential functions other than A(t) =Upper A 0 e Superscript kt. For example, if an i

laila [671]

The question is incomplete. Here is the complete question.

Uninhibited growth can be modeled by exponential functions other than $A(t)=A_{0}e^{kt}$ . for example, if an initial population P₀ requires n units of time to triple, then the function $P(t)=P_{0}(3)^{\frac{t}{n} }$ models the size of the population at time t. An insect population grows exponentially. Complete the parts a through d below.

a) If the population triples in 30 days, and 50 insects are present initially, write an exponential function of the form $P(t)=P_{0}(3)^{\frac{t}{n} }$ that models the population.

b) What will the population be in 47 days?

c) When wil the population reach 750?

d) Express the model from part (a) in the form $A(t)=A_{0}e^{kt}$ .

Answer: a) $P(t)=50(3)^{\frac{t}{30} }$

b) P(t) = 280 insects

c) t = 74 days

d) $A(t)=50e^{0.037t}$

Step-by-step explanation:

a) n is time necessary to triple the population of insects, i.e., n = 30 and P₀ = 50. So, Exponential equation for growth is

$P(t)=50(3)^{\frac{t}{30} }$

b) In t = 47 days:

$P(t)=50(3)^{\frac{t}{30} }$

$P(47)=50(3)^{\frac{47}{30} }$

$P(47)=50(3)^{1.567}$

P(47) = 280

In 47 days, population of insects will be 280

c) P(t) = 750

$750=50(3)^{\frac{t}{30} }$

$\frac{750}{50}=(3)^{\frac{t}{30} }$

$(3)^{\frac{t}{n} }=15$

Using the property <u>Power</u> <u>Rule</u> of logarithm:

$log(3)^{\frac{t}{30} }=log15$

$\frac{t}{30}log(3)=log15$

$t=\frac{log15}{log3} .30$

t = 74

To reach a population of 750 insects, it will take 74 days

d) To express the population growth into the described form, determine the constant k, using the following:

A(t) = 3A₀ and t = 30

$A(t)=A_{0}e^{kt}$

$3A_{0}=A_{0}e^{30k}$

$3=e^{30k}$

Use Power Rule again:

$ln3=ln(e^{30k})$

$ln3=30k$

$k=\frac{ln3}{30}$

k = 0.037

Equation for exponential growth will be:

$A(t)=50e^{0.037t}$

3 0

3 years ago

What is 4.75 as an improper fraction in simplest form

jek_recluse [69]

4.75 = 4 75/100 =
= 475/100 = ( we can divide numerator and denominator by 25 ) = 19/4
Answer:
4.75 as an improper fraction in simplest form is 19/4.

8 0

2 years ago

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5. A study found that the average time it took a person to find their dream home was 5.9 months. If a sample of

solong [7]

Answer:

The 95% confidence interval of the mean time it took a person to find their dream home is between 5.64 months and 6.16 months.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96\frac{0.8}{\sqrt{36}} = 0.26$

The lower end of the interval is the sample mean subtracted by M. So it is 5.9 - 0.26 = 5.64 months

The upper end of the interval is the sample mean added to M. So it is 5.9 + 0.26 = 6.16 months.

The 95% confidence interval of the mean time it took a person to find their dream home is between 5.64 months and 6.16 months.

4 0

3 years ago