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ZanzabumX [31]
3 years ago
12

HEKP PLEASE GOAIDVNL W{DKAIPNV A?? ?F"AW"F"AW FAOW\

Mathematics
1 answer:
Aneli [31]3 years ago
7 0

Answer:

a hope it helps make brainlliest ty

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6h^5+0h^4-12h^3+0h^3+0h=0
aliya0001 [1]
<h2>The value of h = 0,  ± \sqrt{2}</h2>

Step-by-step explanation:

We have,

6h^5+0h^4-12h^3+0h^3+0h=0

To find, the value of h = ?

∴ 6h^5+0h^4-12h^3+0h^3+0h=0

⇒ 6h^5 + (0)h^4 - 12h^3 + (0)h^3 + (0)h = 0

⇒ 6h^5 + 0 - 12h^3 + 0 + 0 = 0

⇒ 6h^5  - 12h^3 = 0

Taking 6 h^3 as common, we get

6h^3(h^2-2) = 0

⇒6h^3 = 0 or,   h^2 - 2 = 0

⇒ 6h^3 = 0 ⇒ h = 0

⇒ h^2 = 2  

⇒ h = ± \sqrt{2}

Hence, the value of h = 0,  ± \sqrt{2}

7 0
3 years ago
Deion misses 4% of the free throws he attempts in a seaon. how many total free throws did he attempt if he missed 17
Burka [1]

He has thrown 425 free throws if he missed 17

<h3>How to determine the total number of free throws?</h3>

The given parameters are:

Proportion of free throw missed, p = 4%

Number of free throw missed, n = 17

Let the total number of throws be N.

So we have

n = p * N

Substitute the known values in the above equation

17 = 4% * N

Divide both sides by 4%

N = 425

Hence, he has thrown 425 free throws if he missed 17

Read more about proportion at:

brainly.com/question/18437927

#SPJ1

3 0
2 years ago
4. The data set DDT (MASS) contains independent measurements of the pesticide DDT on kale. Make a histogram and a boxplot of the
padilas [110]

Answer:

Step-by-step explanation:

Rcode:

library(MASS)

data(DDT,package="MASS")

head(DDT)

colnames(DDT)

print(DDT)

boxplot(DDT)

hist(DDT,main="Histogram for DDT")

fivenum(DDT)

dim(DDT)

outlier_values <- boxplot.stats(DDT)$out # outlier values.

boxplot(DDT, main="DDT", boxwex=0.1)

mtext(paste("Outliers: ", paste(outlier_values, collapse=", ")), cex=0.6)

mean(DDT)

sd(DDT)

Output: * see attachment below*

From Histogram the mean is at the peak

it lies in between 3and 3.5

Form Boxplot we found one outlier and it is 4.64

Fivenum summary is

2.790 3.075 3.220 3.360 4.640

min=2.790

Q1=3.075

Q2=3.220

Q3=3.360

max=4.640

From boxplot the center line is median which cuts the box into 2 equal halves

Distribution is symmetrical

Follows normal distribution.

Mean=3.328

Standard deviation=0.4371531

8 0
3 years ago
What is the answer to the equation 1 + 1​
Ivahew [28]

Answer:

11

21

or.....

2?!?!

Step-by-step explanation:

!!!!!!!!!

8 0
3 years ago
PLS HELP ME FAST PLS !!!!!!!!!!!
erastova [34]

Answer:

The Answer is P = (-2.5 , 1 )

Step-by-step explanation:

Let, the point is P(x,y)

then, x=\frac{m_{1}x_{2}+ m_{2}x_{1}}{m_{1}+m_{2}}

     =>x=\frac{5*(-1)+ 3*(-5)}{5+3}

     ∴x=-2.5

Again, y=\frac{m_{1}y_{2}+ m_{2}y_{1}}{m_{1}+m_{2}}

     =>y=\frac{5*7+ 3*(-9)}{5+3}

     ∴y=1

Thus, P = (-2.5 , 1 )

6 0
3 years ago
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