What is the value of the expression ?

Mathematics

1 answer:

kherson [118]3 years ago

6 0

<h2> $16 - 3(8 - 3) {}^{2} \div 5$ </h2><h2>ANSWER</h2>

<h2> $1$ </h2><h3>EXPLANATION</h3>

$16 - 3 \times 5 {}^{2} \div 5$

$16 - 3 \times 5 {}^{2} \div 5 {}^{1}$

$16 - 3 \times 5 {}^{2} \div 5 {}^{ - 1}$

$16 - 3 \times 5 {}^{2 - 1}$

$16 - 3 \times 5 {}^{1}$

$16 - 3 \times 5$

$16 - 15$

$1$

HopeItHelps you

#CarryOnLearning

You might be interested in

A container contains 1212 diesel engines. The company chooses 88 engines at random, and will not ship the container if any of t

bija089 [108]

Answer:

The probability that a container will be shipped even though it contains 2 defectives if the sample size is 88, will be $P(S)=85.99\%$

Step-by-step explanation:

The first step is to count the number of total possible random sets of taking a sample size of 88 engines over 1212 engines of the population, so $\left[\begin{array}{ccc}1212\\88\end{array}\right] =1212C88=4.7205x10^{135}$

The second step is to count the number of total possible random sets of taking a sample size of 88 engines over 1210 engines (discounting the 2 defective engines) as the possible ways to succeed, so $\left[\begin{array}{ccc}1210\\88\end{array}\right] =1212C88=4.0596x10^{135}$

Finally we need to compute $\frac{\# ways\ to\ succeed}{\# random\ sets\ of \ 88} =\frac{4.0596x10^{135}}{4.7205x10^{135}}=0.8599=P(S)$ , therefore the probability that a container will be shipped is $P(S)=85.99\%$