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3 years ago

Help me find out this prom or

Mathematics

2 answers:

Firlakuza [10]3 years ago

4 0

Answer:

80 percent

Step-by-step explanation:

Talja [164]3 years ago

4 0

Answer:

80 percent

Step-by-step explanation:

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Explain why the polynomial 144x2 − 200x + 81 is not a perfect square.

Sloan [31]

The square root of 144x^2 = 12 x
and square root of + 81 = 9 or -9

if this polynomial were a perfect square then its factors would be
(12x - 9)(12x - 9)

but its not a perfect squre because (12x - 9)(12x - 9) = 144x^2 - 216x + 81.

5 0

3 years ago

Fill in the y-value for each box in the t-chart using the equation below !?

riadik2000 [5.3K]

3,4,7 just plug the x into the equation

3 0

3 years ago

4(x-6)<-2x+6 what is the solution to the inequality _

Cloud [144]

Simplify both sides if needed. The left-hand side needs simplification.

4(x - 6) $\leq$ -2x + 6
4x - 24 $\leq$ -2x + 6

All is left to do is add and subtract to get the x variable all alone.

4x - 24 $\leq$ -2x + 6
6x - 24 $\leq$ 6 <-- Add 2x to both sides
6x $\leq$ 30 <-- Add 24 to both sides
x $\leq$ 5 <-- Divide both sides by 6

In order to be in the solution set, x has to be less than or equal to 5.

In interval notation: [5, -∞)

7 0

3 years ago

Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally

rusak2 [61]

Answer:

(a) The probability that the thickness is less than 3.0 mm is 0.119.

(b) The probability that the thickness is more than 7.0 mm is 0.119.

(c) The probability that the thickness is between 3.0 mm and 7.0 mm is 0.762.

Step-by-step explanation:

We are given that thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed, with a mean of 5.0 millimeters (mm) and a standard deviation of 1.7 mm.

Let X = thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village.

So, X ~ Normal( $\mu=5.0,\sigma^{2} =1.7^{2}$ )

The z-score probability distribution for the normal distribution is given by;

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

where, $\mu$ = population mean thickness = 5.0 mm

$\sigma$ = standard deviation = 1.7 mm

(a) The probability that the thickness is less than 3.0 mm is given by = P(X < 3.0 mm)

P(X < 3.0 mm) = P( $\frac{X-\mu}{\sigma}$ < $\frac{3.0-5.0}{1.7}$ ) = P(Z < -1.18) = 1 - P(Z $\leq$ 1.18)

= 1 - 0.8810 = 0.119

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

(b) The probability that the thickness is more than 7.0 mm is given by = P(X > 7.0 mm)

P(X > 7.0 mm) = P( $\frac{X-\mu}{\sigma}$ > $\frac{7.0-5.0}{1.7}$ ) = P(Z > 1.18) = 1 - P(Z $\leq$ 1.18)

= 1 - 0.8810 = 0.119

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

(c) The probability that the thickness is between 3.0 mm and 7.0 mm is given by = P(3.0 mm < X < 7.0 mm) = P(X < 7.0 mm) - P(X $\leq$ 3.0 mm)

P(X < 7.0 mm) = P( $\frac{X-\mu}{\sigma}$ < $\frac{7.0-5.0}{1.7}$ ) = P(Z < 1.18) = 0.881

P(X $\leq$ 3.0 mm) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{3.0-5.0}{1.7}$ ) = P(Z $\leq$ -1.18) = 1 - P(Z < 1.18)

= 1 - 0.8810 = 0.119

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

Therefore, P(3.0 mm < X < 7.0 mm) = 0.881 - 0.119 = 0.762.

4 0

3 years ago

Solve for ppp. 16-3p=\dfrac23p+516−3p= 3 2 p+516, minus, 3, p, equals, start fraction, 2, divided by, 3, end fraction, p, plus

Ksivusya [100]

Given:

The given equation is:

$16-3p=\dfrac{2}{3}p+5$

To find:

The value of p.

Solution:

We have,

$16-3p=\dfrac{2}{3}p+5$

Multiply both sides by 3.

$3(16-3p)=3\left(\dfrac{2}{3}p+5\right)$

$48-9p=2p+15$

Isolating the variable terms, we get

$48-15=2p+9p$

$33=11p$

Divide both sides by 11, we get

$\dfrac{33}{11}=p$

$3=p$

Therefore, the required solution is $p=3$ .

7 0

3 years ago