Please answer If I get this right I'm done for the week I will give brainiest if you're right!!!

Mathematics

2 answers:

Evgen [1.6K]3 years ago

6 0

Answer:

they are not equivilent

Step-by-step explanation:

Hatshy [7]3 years ago

4 0

Answer:

The answer is true.

Step-by-step explanation:

$\frac{4}{6} = \frac{14}{21} \\ 84 = 84$

Because the last equation, 84=84, is true, then the ratios are true.

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Which number is greater 17×10^8 4×10^8

Sunny_sXe [5.5K]

Answer:

17×10^8 is greater

Step-by-step explanation:

The expontent comes first which gives you

1000000000 then you multiply that with 17 to get

1700000000

3 0

3 years ago

Read 2 more answers

2 more and I’m done thank y’all so much sorry I’m dumb and in 8th grade LOL

kotegsom [21]

9. -5(4+6r)

-20-30r

-20-30(4)=-20-120=-140

10. -9-n/3

-9-9/3=-9-3=-12

answers: -140, -12

7 0

3 years ago

Read 2 more answers

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:

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

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