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3 years ago

HELP AGAIN TRUST ME ITS A LOT! BRAINLIEST!!

Mathematics

1 answer:

Ksju [112]3 years ago

5 0

Answer:

they lost 6 yards then 3 yards 6+3=9 so they behind the line of scrimmage by 9 yards

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If X = 7 units, Y = 4 units, Z = 18 units, and h = 7 units, what is the surface area of the triangular prism shown above?

nata0808 [166]

Answer:

Step-by-step explanation:

3 0

3 years ago

Determine the decibel values of thermal noise power (N) and thermal noise power density (No) given the following: T=292 degrees

Agata [3.3K]

Answer: d. None of the above are correct.

Step-by-step explanation: Noise is a superfluous random alteration in an eletrical signal. There are different types of noises created by different devices and process. Thermal noise is one of them. It is unavoidable because is created by the agitation of the charge carriers, due to temperature, inside an eletrical conductor at equilibrium and is present in all eletrical circuits.

The formula to find the thermal noise power (N) is: N = $k_{b}$ .T.B, where:

$k_{b}$ is Boltzmann constant (1.38. $10^{-23}$ J/K);

T is temperature in Kelvin;

B is the bandwith;

Calculating the thermal noise power:

N = 1.38. $10^{-23}$ ·292·40

N = 16118.4. $10^{-23}$ dBm

The thermal noise power [N] = 16118.4. $10^{-23}$ dBm

Noise power density or simply Noise density (N₀) is the noise power per unit of bandwith and its SI is watts per hertz.

For thermal noise, N₀ = kT, where

<em>k </em>is the Boltzmann constant in J/K;

T is the receiver system noise temperature in K;

N₀ = 1.38. $10^{-23}$ . 292

N₀ = 402.96. $10^{-23}$ W/Hz

The thermal noise power density [N₀] = 402.96. $10^{-23}$ W/Hz

7 0

3 years ago

Explain in words,how to find the product of -4(-1.5) using a number line. Where do you end up?

Gemiola [76]

Answer:

so u multiply negative four times nagative 1.5

6 0

3 years ago

Construct the confidence interval for the population standard deviation for the given values. Round your answers to one decimal

Alex777 [14]

Answer:

The correct answer is " $2.633< \sigma < 4.480$ ".

Step-by-step explanation:

Given:

n = 21

s = 3.3

c = 0.9

now,

$df = n-1$

$=20$

⇒ $x^2_{\frac{\alpha}{2}, n-1 }$ = $x^2_{\frac{0.9}{2}, 21-1 }$

= $31.410$

⇒ $x^2_{1-\frac{\alpha}{2}, n-1 }$ = $10.851$

hence,

The 90% Confidence interval will be:

= $\sqrt{\frac{(n-1)s^2}{x^2_{\frac{\alpha}{2}, n-1 }} } < \sigma < \sqrt{\frac{(n-1)s^2}{x^2_{1-\frac{\alpha}{2}, n-1 }}$

= $\sqrt{\frac{(21-1)3.3^2}{31.410} } < \sigma < \sqrt{\frac{(21.1)3.3^2}{10.851} }$

= $\sqrt{\frac{20\times 3.3^2}{31.410} } < \sigma < \sqrt{\frac{20\times 3.3^2}{10.851} }$

= $2.633< \sigma < 4.480$

4 0

2 years ago

35 solve the following syste

blondinia [14]

Part A: The solution is $(-0.923,5.692)$

Part B: The point $(3,7)$ is not in the solution set.

Explanation:

Part A: The given inequalities are $3 x+4 y>20$ and $x$

The solution can be determined by solving the two inequalities by substitution method.

Changing inequalities to equality, we have,

$x=3 y-18$ and $3 x+4 y=20$

Let us substitute $x=3 y-18$ in the equation $3 x+4 y=20$ , we get,

$3 (3y-18)+4 y=20$

$9y-54+4y=20$

$13y=74$

$y=5.692$

Substituting $y=5.692$ in $x=3 y-18$ , we get,

$x=3 (5.692)-18$

$=17.076-18$

$x=-0.923$

Thus, the solution set is $(-0.923,5.692)$

Part B: Now, we shall determine whether the point $(3,7)$ is in the solution set.

Let us substitute the point $(3,7)$ in the inequalities $3 x+4 y>20$ and $x$ , we get,

$3 (3)+4 (7)>20$

$9+28>20$

$37>20$

Also, substituting $(3,7)$ in $x$ , we get,

$3$

Since, the point $(3,7)$ does not satisfy one of the inequality $x$ , the solution set does not contain the point $(3,7)$

Thus, the point $(3,7)$ is not in the solution set.

8 0

3 years ago