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arsen [322]
3 years ago
15

PLEASE HELP! I WILL MARK BRAINLIEST

Mathematics
1 answer:
bogdanovich [222]3 years ago
7 0

9514 1404 393

Answer:

  1.  4463

  2. no

Step-by-step explanation:

1. Put the value of x into the equation and do the arithmetic.

  f(4) = 5000(0.972^4) ≈ 5000·0.8926168 ≈ 4463.08

There will be about 4463 lemmings in 4 years.

__

2. An exponential function has the variable in the exponent. When the variable is the base of a term that has an integer constant exponent, we call it a polynomial function.

The function is NOT an exponential function.

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Step-by-step explanation:

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What is the minimum value of 2x + 2y in the feasible region?<br> (0,7) (3,7) (6,3) (6,0)
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Graph the feasible region for the following constraints: x + y < 5. 2x + y > 4 ... y = 10/3. x = 30/3 - 10/3 = 20/3. Intersects at (20/3, 10/3). -x + 2y = 0. x - 2y = 0.
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Step-by-step explanation:

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3 years ago
The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e
I am Lyosha [343]

Answer:

(a) The proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b) The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

Step-by-step explanation:

Let <em>X</em> = number of students who read above the eighth grade level.

(a)

A sample of <em>n</em> = 269 students are selected. Of these 269 students, <em>X</em> = 224 students who can read above the eighth grade level.

Compute the proportion of students who can read above the eighth grade level as follows:

\hat p=\frac{X}{n}=\frac{224}{269}=0.8327

The proportion of students who can read above the eighth grade level is 0.8327.

Compute the proportion of tenth graders reading at or below the eighth grade level as follows:

1-\hat p=1-0.8327

        =0.1673

Thus, the proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b)

the information provided is:

<em>n</em> = 709

<em>X</em> = 546

Compute the sample proportion of tenth graders reading at or below the eighth grade level as follows:

\hat q=1-\hat p

  =1-\frac{X}{n}

  =1-\frac{546}{709}

  =0.2299\\\approx 0.229

The critical value of <em>z</em> for 95% confidence interval is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Compute the 95% confidence interval for the population proportion as follows:

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

     =0.229\pm 1.96\times \sqrt{\frac{0.229(1-0.229)}{709}}\\=0.229\pm 0.03136\\=(0.19764, 0.26036)\\\approx (0.198, 0.260)

Thus, the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

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3 years ago
A circular pan has a diameter of 8 inches. What is the exact area of the pan?
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Read 2 more answers
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