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2 years ago

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"A day care program has an average daily expense of $75.00. The standard deviation is $5.00. The owner takes a sample of 64 bill

s. What is the probability the mean of his sample will be between $70.00 and $80.00?"

2 answers:

kolbaska11 [484]2 years ago

8 0

Answer:

p=1

Step-by-step explanation:

Given that a day care program has an average daily expense of $75.00. The standard deviation is $5.00.

Sample size = 64

Std error of sample = $\frac{5}{\sqrt{64} } \\=0.625$

Thus sample mean will follow normal with

(75, 0.625)

Required probability

= $P(70$

It is almost a certain event with p =1

kirill115 [55]2 years ago

4 0

Answer:

their you go i had it wrong so thought i give someone the right answer

Step-by-step explanation:

1.0

34

1.0

34

68

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Read 2 more answers

The equation p(t) = 1.e represents a

zvonat [6]

Answer:

(b), (d) and (e)

Step-by-step explanation:

Given

$p(t) = 1 * e^t$

See attachment for $y = p(t)$

Required

Select true statements from the given options

(a) $\ln(30)$ = days the bacteria reaches 30000

We have:

$p(t) = 1 * e^t$

In this case:

$t = \ln(30)$ and $p(t) = 30000$

So, we have:

$30000 = 1 * e^{\ln(30)}$

$30000 = e^{\ln(30)}$

Using a calculator, we have:

$e^{\ln(30)} = 30$

So:

$30000 = 30$

The above equation is false.

(a) is not true

(b) The graph shows that $\ln(20) \approx 3$

We have:

$p(t) = 1 * e^t$

Let t = 3

So;

$p(3) = 1 * e^3$

From the graph, $p(3) = 20$

So:

$20 = 1 * e^3$

$20 = e^3$

Take natural logarithm of both sides

$\ln(20) = \ln(e^3)$

This gives:

$\ln(20) = 3$

(b) is true

(c) $\ln(t) = y$ is the logarithm form of $y = e^t$

We have:

$y = e^t$

Take natural logarithm of both sides

$\ln(y) = \ln(e^t)$

This gives:

$\ln(y) = t$

$\ln(y) = t$ $\ne$ $\ln(t) = y$

(c) is false

(d) $e^4 > 50$ and $\ln(50) < 4$

From the graph, we have:

$e^4 = 54$ --- rough readings

This implies that:

$e^4 > 50$ is true

Because $54 > 50$

Take natural logarithm of both sides

$\ln(54) > \ln(50)$

Rewrite as:

$\ln(50) < \ln(54)$

We have:

$e^4 = 54$

Take natural logarithm of both sides

$\ln(e^4) = \ln(54)$

$4 = \ln(54)$

$\ln(54) = 4$

Substitute $\ln(54) = 4$ in $\ln(50) < \ln(54)$

$\ln(50) < 4$

(d) is true

(e) The graph shows that $10 \approx \ln(2.3)$

We have:

$p(t) = 1 * e^t$

Let t = 2.3

So;

$p(2.3) = 1 * e^{2.3}$

From the graph,

$p(2.3) = 10$ ---- rough readings

So:

$10 = 1 * e^{2.3}$

$10 = e^{2.3}$

Take natural logarithm of both sides

$\ln(10) = \ln(e^{2.3})$

This gives:

$\ln(10) = 2.3$

(e) is true

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