I REALLY NEED THIS NOW. PLEASE HELP ME!!

What is the area? Bit confused on this one.

Nutka1998 [239]

Answer:

A = 9x² + 12x

Step-by-step explanation:

The area (A) of the rectangle is calculated as

A = height × width

= 3(3x² + 4x) ← distribute parenthesis by 3

= 9x² + 12x

7 0

3 years ago

Read 2 more answers

[297÷3 + { 4 × 3 ( 20 – 36 ÷ 5 – 2)}]

maw [93]

Here is the answer to your problem:

<h3><u>https://brainly.in/question/10886862</u></h3>

4 0

3 years ago

What is th he sum of the first seven terms of the series -3+6-12+24-...?

ohaa [14]

Answer:

-129

Step-by-step explanation:

-3+6-12+24-48+96-192=

6+24+96-3-12-48-192=

126-255=

-129

Hope this helps!

5 0

3 years ago

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.

nika2105 [10]

Answer:

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

$\bar X=2.55$ represent the sample mean for the late time for a flight

$\mu$ population mean

$\sigma=12$ represent the population deviation

n=76 represent the sample size

Confidence interval

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

The confidence interval for the true mean is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The Confidence level given is 0.95 or 95%, th significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ . If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got $z_{\alpha/2}=1.96$

Replacing we got:

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0

3 years ago

1) The deer population in a national park is expected to decline over time. The park

Lorico [155]

Answer:

Here's what I find.

Step-by-step explanation:

You have 800 deer at the end of Year 1, and you expect the population to decrease each year thereafter.

a) i) The recursive formula

Let dₙ = the deer population n years after the initial measurement.

$d_{n} = d_{n - 1}r^{n}$

For this situation,

$d_{n} = d_{n - 1}(0.5)^{n}$

a) ii) Definitions

n = the number of years from first measurement

r = the common ratio, that is, the deer population at the end of one year divided by the population of the previous year.

a) iii) First term of sequence

The first term of the sequence is d₀, the population when first measured.

b) The function formula

The formula for the nth term of a geometric series is

$d_{n} =d_{0}r^{n - 1}$

c) Value of d₀

Let n = 2; then d₂ = 800

$\begin{array}{rcl}800 & = & d_{0}(0.5)^{2 - 1}\\800 & = & d_{0}(0.5)\\\\d_{0} & = & \dfrac{800}{0.5}\\\\& =&\mathbf{1600}\\\end{array}$

4 0

3 years ago