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3 years ago

Which functions are decreasing? Select all answers that are correct.

Mathematics

2 answers:

bazaltina [42]3 years ago

8 0

Answer:

it's number 3 and number 4

i took the quiz

monitta3 years ago

7 0

Answer:

The last two functions are decreasing!

Step-by-step explanation:

You can easily tell the functions are decreasing if the function is going down from L to R. If it is going up, from L to R, then it is increasing!

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48+(2*13)/110= ?<br> A) .673<br> B) .7214<br> C) 51.5<br> D) 35.62

siniylev [52]

48 + (2*13) / 110
PE(MD)(AS)

2 * 13 = 26
48 + 26 / 110

26 / 110 = 0.236..
Rounded = 0.24

48 + 0.24 = 48.24….

??????…

6 0

2 years ago

Find the area of a sector of a circle with radius 6 centimetre if angle of sector is 60 degree

Gnesinka [82]

Answer:

Step-by-step explanation:

$area=\frac{\pi r^2 \times\theta}{360} =\frac{\pi 6^2 \times 60}{360} =6 \pi \approx 18.85~ cm^2$

6 0

3 years ago

Barb is making a necklace. She string 1 white bead,then 3 blue beads,then 1 white bead, and so on. Write the numbers for the fir

KiRa [710]

There will be 3 white beads out of the first 8.

8 0

3 years ago

Read 2 more answers

A flat rectangular piece of aluminum has a perimeter of 64 inches. The length is 8 inches longer than the width. Find the width.

denpristay [2]

Answer:

w = 12

Step-by-step explanation:

W=48/4

W=12 ANS. FOR THE WIDTH.

L=12+8=20 ANS. FOR THE LENGTH.

PROOF:

2*20+2*12=64

40+24=64

64=64

7 0

3 years ago

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65

erma4kov [3.2K]

Answer:

Probability that the sample average is at most 3.00 = 0.98030

Probability that the sample average is between 2.65 and 3.00 = 0.4803

Step-by-step explanation:

We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85.

Also, a random sample of 25 specimens is selected.

Let X bar = Sample average sediment density

The z score probability distribution for sample average is given by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean = 2.65

$\sigma$ = standard deviation = 0.85

n = sample size = 25

(a) Probability that the sample average sediment density is at most 3.00 is given by = P( X bar <= 3.00)

P(X bar <= 3) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ <= $\frac{3-2.65}{\frac{0.85}{\sqrt{25} } }$ ) = P(Z <= 2.06) = 0.98030

(b) Probability that sample average sediment density is between 2.65 and 3.00 is given by = P(2.65 < X bar < 3.00) = P(X bar < 3) - P(X bar <= 2.65)

P(X bar < 3) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{3-2.65}{\frac{0.85}{\sqrt{25} } }$ ) = P(Z < 2.06) = 0.98030

P(X bar <= 2.65) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ <= $\frac{2.65-2.65}{\frac{0.85}{\sqrt{25} } }$ ) = P(Z <= 0) = 0.5

Therefore, P(2.65 < X bar < 3) = 0.98030 - 0.5 = 0.4803 .

8 0

4 years ago