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2 years ago

GEOMETRY please Help

Mathematics

1 answer:

geniusboy [140]2 years ago

7 0

Answer:

A right circular cone

Step-by-step explanation:

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g red bell pepper seeds germinates 85% of the time. planted 25 seeds. What is the probability that 20 or more germinate

bixtya [17]

Answer:

$P(X\geq 20)= P(X=20)+P(X=21)+P(X=22)+P(X=23)+P(X=24)+P(X=25)$

And replacing using the mass function we got:

$P(X=20)=(25C20)(0.85)^{20} (1-0.85)^{25-20}=0.156$

$P(X=21)=(25C21)(0.85)^{21} (1-0.85)^{25-21}=0.211$

$P(X=22)=(25C22)(0.85)^{22} (1-0.85)^{25-22}=0.217$

$P(X=23)=(25C23)(0.85)^{23} (1-0.85)^{25-23}=0.161$

$P(X=24)=(25C24)(0.85)^{24} (1-0.85)^{25-24}=0.0759$

$P(X=25)=(25C25)(0.85)^{25} (1-0.85)^{25-25}=0.0172$

And adding the values we got:

$P(X\geq 20) = 0.8381$

Step-by-step explanation:

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=25, p=0.85)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

We want to find the following probability:

$P(X\geq 20)= P(X=20)+P(X=21)+P(X=22)+P(X=23)+P(X=24)+P(X=25)$

And replacing using the mass function we got:

$P(X=20)=(25C20)(0.85)^{20} (1-0.85)^{25-20}=0.156$

$P(X=21)=(25C21)(0.85)^{21} (1-0.85)^{25-21}=0.211$

$P(X=22)=(25C22)(0.85)^{22} (1-0.85)^{25-22}=0.217$

$P(X=23)=(25C23)(0.85)^{23} (1-0.85)^{25-23}=0.161$

$P(X=24)=(25C24)(0.85)^{24} (1-0.85)^{25-24}=0.0759$

$P(X=25)=(25C25)(0.85)^{25} (1-0.85)^{25-25}=0.0172$

And adding the values we got:

$P(X\geq 20) = 0.8381$

7 0

3 years ago

Help please? 10 points. Correctly

Romashka [77]

From the Figure :

Point A is (-3 , -3)

Point B is (6 , 6)

We know that, The Mid Point of Two Points (x₁ , y₁) and (x₂ , y₂) is given by :

$\mathsf{\implies [(\frac{x_1 + x_2}{2})\;,\;(\frac{y_1 + y_2}{2})]}$

$\mathsf{\implies Midpoint\;of\;Line\;AB\;is\;[(\frac{-3 + 6}{2})\;,\;(\frac{-3 + 6}{2})]}$

$\mathsf{\implies Midpoint\;of\;Line\;AB\;is\;[(\frac{3}{2})\;,\;(\frac{3}{2})]}$

$\mathsf{\implies Midpoint\;of\;Line\;AB\;is\;(1.5,1.5)}$

4 0

3 years ago

Joseph claims that a scatterplot in which the y-values increase as the x-values increase must have a linear association. Amy cla

quester [9]

Answer:

Amy is correct because a nonlinear association could increase along the whole data set, while being steeper in some parts than others. The scatterplot could be linear or nonlinear.

Step-by-step explanation:

I just took the quiz, I hope this helps :).

7 0

3 years ago

Read 2 more answers

Linda received a credit card through the mail. The interest rate charged by the credit card company is 24.9%. If she maintains a

creativ13 [48]

<span>i think the answer is $597.60 </span>

6 0

3 years ago

5. The area of the trapezoid is 75 square inches. Find the height of the trapezoid. And it’s two bases are 10 and 15.

GaryK [48]

6

6
h/2(sum of bases) =75=area
h=3*2

Mark brainliest please

7 0

2 years ago