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3 years ago

Will give brainliest! 30 points per answer!

Mathematics

1 answer:

-Dominant- [34]3 years ago

5 0

Answer:

1) Yes, there are many transformation present in the conductor.

2) Transformation present in the construction is in the front side of it. there are many transformation such as reflections or rotation. it can rotate .

A transformation is a process that manipulates a polygon or other two-dimensional object on a plane or coordinate system. Mathematical transformations describe how two-dimensional figures move around a plane or coordinate.

In a geometric figures there are 4 types of Transformation-

i) Translations - Is the construction can stand on an axis like earth.

ii) Reflection

iii) Rotation

iv) Dilation

Pur construction is of rigid transformation means it does not change the shape of construction .

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49/10 as a mixed number

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⇒ $\frac{49}{10}$

$\frac{49}{10}$ is a improper fraction.
$\frac{49}{10}$ as a mixed number would be 4 $\frac{9}{10}$

⇒10 x 4 = 40
⇒40 + 9 = 49

$\frac{49}{10}$ = 4 $\frac{9}{10}$

⇒ANSWER = 4 $\frac{9}{10}$

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3 years ago

What should I buy? A study conducted by a research group in a recent year reported that of cell phone owners used their phones i

Llana [10]

Answer:

The probability that seven or more of them used their phones for guidance on purchasing decisions is 0.7886.

Step-by-step explanation:

The question is incomplete:

What should I buy? A study conducted by a research group in a recent year reported that 57% of cell phone owners used their phones inside a store for guidance on purchasing decisions. A sample of 14 cell phone owners is studied. Round the answers to at least four decimal places. What is the probability that seven or more of them used their phones for guidance on purchasing decisions?

We can model this as a binomial random variable, with p=0.57 and n=14.

$P(x=k)=\dbinom{n}{k} p^{k}q^{n-k}$

a) We have to calculate the probability that seven or more of them used their phones for guidance on purchasing decisions:

$P(x\geq7)=\sum_{k=7}^{14}P(x=k)\\\\\\$

$P(x=7)=\dbinom{14}{7} p^{7}q^{7}=3432*0.0195*0.0027=0.1824\\\\\\P(x=8) = \dbinom{14}{8} p^{8}q^{6}=3003*0.0111*0.0063=0.2115\\\\\\P(x=9) = \dbinom{14}{9} p^{9}q^{5}=2002*0.0064*0.0147=0.1869\\\\\\P(x=10) = \dbinom{14}{10} p^{10}q^{4}=1001*0.0036*0.0342=0.1239\\\\\\P(x=11) = \dbinom{14}{11} p^{11}q^{3}=364*0.0021*0.0795=0.0597\\\\\\P(x=12) = \dbinom{14}{12} p^{12}q^{2}=91*0.0012*0.1849=0.0198\\\\\\P(x=13) = \dbinom{14}{13} p^{13}q^{1}=14*0.0007*0.43=0.004\\\\\\$

$P(x=14) = \dbinom{14}{14} p^{14}q^{0}=1*0.0004*1=0.0004\\\\\\$

$P(x\geq7)=0.1824+0.2115+0.1869+0.1239+0.0597+0.0198+0.004+0.0004\\\\P(x\geq7)=0.7886$

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