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3 years ago

Sandra has a 1/3 chance of making the team. What is the probability that Sandra will not make the team?

Mathematics

2 answers:

NemiM [27]3 years ago

8 0

Answer:

2/3 is the probability that Sandra will not make the team.

Hope it helps!!!

babymother [125]3 years ago

4 0

Answer:

2/3

Step-by-step explanation:

If something will happen with 100% certainty, then the probability is 1 or 100%.

If she has a 1/3 chance of making the team, the the chance of not making the team is 1 - 1/3 which equals 2/3.

Answer: 2/3

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Find the cost of painting a cubical block of side length 3.2 cm, if painting costs $5 per cm2.

SashulF [63]

<h3><u>Answer</u>:- </h3>

$\longrightarrow \sf \$ \: 307.2$

<h3><u>Solution:-</u></h3>

Since painting is done on outer surface of cubical block we will have to find its surface area which is given by-

$\green{ \underline { \boxed{ \sf{Surface \: area \:of \:Cube = 6a^2}}}}$

<u>where a is side of the cube </u>

$\begin{gathered}\\\quad \sf Surface \: area \:of \:Cube = 6\times(3.2)^2 \\\end{gathered}$

$\begin{gathered}\\\implies \quad \sf 6\times 3.2\times3.2 \\\end{gathered}$

$\begin{gathered}\\\implies \quad \sf 61.44 cm^2 \\\end{gathered}$

Now,

$\maltese \sf Cost \:of \:painting \: 1 \: cm² = \$ 5$

$\sf Cost \: of \: painting \: 61.44 \:cm²=61.44×5$

$\sf\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad=\$ 307.2$

$\leadsto$ Thus it would cost $ 307.2 to paint the entire cubical block.

3 0

2 years ago

What is the absolute value of | 56-12 |?

muminat

Answer: 44! Hope this helps :D

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

If l=10, b=5, h=2, find the values of 2h(l+b)

Eddi Din [679]

Answer:

Step-by-step explanation:

given:

l = 10

b = 5

h = 2

to find:

2h(l + b)

substitute the given values of l , b and h

=2*2(10 + 5)

=4*15

=60

3 0

3 years ago

Read 2 more answers

A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the

Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average age = 25 years

$\sigma$ = population standard deviation = 1.8 years

n = sample of students = 36

$\mu$ = population average age

So, 98% confidence interval for the average age, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < ${\bar X - \mu}$ < $2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

P( $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ , $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ]

= [ $25 - 2.3263 \times {\frac{1.8}{\sqrt{36} }$ , $25 + 2.3263 \times {\frac{1.8}{\sqrt{36} }$ ]

= [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

8 0

3 years ago

What is the value of variable for -4(y - 2) = 12<br> Explain please

Scrat [10]

Answer:

-4(y - 2) = 12

Standard form:

−4y − 4 = 0

Factorization:

−4(y + 1) = 0

Solutions:

y = −4

= −1

Step-by-step explanation:

7 0

3 years ago