How to write 1.06 in expanded form

Mathematics

2 answers:

Lady bird [3.3K]1 year ago

8 0

$\large{\sf{Question:}} \\ \\$

How to write 1.06 in expanded form

Solution :

1.06

$\implies$ 1 + 0.0 + 0.06

zloy xaker [14]1 year ago

7 0

Answer:

1+(6*.01)

Step-by-step explanation:

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Jean sold cupcakes and cookies yesterday. Each cupcake sold for $2.25 and each cookie sold for $0.50. At the end of the day, Jea

slega [8]

Answer:number of cupcakes sold is 13

Number of cookies sold is 27

Step-by-step explanation:

Let x represent the number of cupcakes that Jane sold.

Let y represent the number of cookies that Jane sold.

Each cupcake sold for $2.25 and each cookie sold for $0.50. At the end of the day, Jean had sold $42.75 worth of cookies and cupcakes.

This means that

2.25x + 0.5y = 42.75 - - - - - - - - - -1

she sold 40 cupcakes and cookies combined, it means that

x + y = 40

Substituting x = 40 - y into equation 1, it becomes

2.25(40 - y) + 0.5y = 42.75

90 - 2.25y + 0.5y = 42.75

- 2.25y + 0.5y = 42.75 - 90

- 1.75y = - 47.25

y = - 47.35/-1.75

y = 27

x = 40 - y

x = 40-27 = 13

4 0

3 years ago

A telemarketer is successful at getting people to donate money for her organization in 55% of all calls she makes. She must get

Tems11 [23]

An interesting twist to a binomial distribution problem.

Given:
p=55%=0.55 for probability of success in solicitation
x=4=number of successful solicitations
n=number of calls to be made
P(x,n,p)>=89.9%=0.899 (from context, it is >= and not =, which is almost impossible)

From context of question, all calls are assumed independent, with constant probability of success, so binomial distribution is applicable.

The number of successes, x, is then given by
$P(x)=C(n,x)p^x(1-p)^{n-x}$ where
p=probability of success
n=number of trials
x=number of successes $C(n,x)=\frac{n!}{x!(n-x)!}$

Here we need n such that
P(x,n,p)>=0.899
given
x>=4, p=0.55, which means we need to find

Method 1: if a cumulative binomial distribution table is available, we can look up n=9,10,11 and find
P(x>=4,9,0.55)=0.834
P(x>=4,10,0.55)=0.898
P(x>=4,11,0.55)=0.939
So she must make (at least) 11 calls to make sure the probability of meeting her quota is 89.9% or more.

Method 2: using technology.
Similar to method 1, we can look up the probabilities directly, for n=9,10,11
P(x>=4,9,0.55)=0.834178
P(x>=4,10,0.55)=0.8980051
P(x>=4,11,0.55)=0.9390368

Method 3: using simple calculator
Here we need to calculate the probabilities for each value of n=10,11 and sum the probabilities of FAILURE S=P(0,n,0.55)+P(1,n,0.55)+P(2,n,0.55)+P(3,n,0.55)
so that the probability of success is 1-S.
For n=10,
P(0,10,0.55)=0.000341
P(1,10,0.55)=0.004162
P(2,10,0.55)=0.022890
P(3,10,0.55)=0.074603
So that
S=0.000341+0.004162+0.022890+0.074603
=0.101995
and Probability of getting 4 successes (or more)
=1-S
=0.898005, missing target by 0.1%

So she will have to make 11 phone calls, bring up the probability to 93.9%. The work is similar to that of n=10.

8 0

3 years ago

Someone please help me I’ll give out brainliest please dont answer if you don’t know

Mashcka [7]

Answer:

6.1 ⋅ $10^{3}$