Which statement is tru about the value of |-5|

Maya reads that 0.63 of Oak School is female and 0.47 of Elm School is female. She says that Oak School must have more females.

-BARSIC- [3]

Answer:

No not correct

Step-by-step explanation:

We are given information that Maya came to know that 0.63 pf Pal school is female and 0.47 of Elm school is female.

Without knowing the total number of students in each school, we cannot conclude that 0.63 is greater than 0.47

For example, say number of students in Oak school be 100, and in Elm school be 200

Females in Oak school = 0.63(100) =63

Females in Elm school = 0.47(200) = 94

We find that there are more females in Elm school

So with the information given, we cannot conclude she is correct , there are chances for her being wrong also.

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3 years ago

Explain how you can use a model 4 x 1/3

Romashka-Z-Leto [24]

You can use it to find the area of something

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3 years ago

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a coffee shop served 24 people before 7:00am. Two-thirds had decaf, and the remaining ordered a specialty coffee. How many custo

Stolb23 [73]

There were 8 customers who ordered specialty coffee. (24/1 x 1/3 = 8)

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2 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

(c) 90.53%

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .