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3 years ago

NEED HELP NOW 100 POINTS

Mathematics

2 answers:

pogonyaev3 years ago

8 0

Answer:

insert is the correct answer hope it helps you

Step-by-step explanation:

kirill [66]3 years ago

3 0

Insert is correct. This is the most logical answer.

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F(-12) if f(x) = 2x + 15

Alenkinab [10]

Answer: $f(-12)=-9$

Step-by-step explanation:

We have the following function:

$f(x)=2x+15$

And we have to evaluate it when $x=-12$ , this means we have to substitute $x$ by $-12$ :

$f(-12)=2(-12)+15$

Solving:

$f(-12)=-24+15$

Finally:

$f(-12)=-9$

8 0

4 years ago

Please help!! I will give brainliest and no links!! Also please do it in a picture model “standard algorithm” please do both que

Alexandra [31]

Answer:

1. 219.571428571

2. 150.314285714

5 0

3 years ago

What is the square root of 13 ?

kipiarov [429]

<span>3.60555128 is the awnser</span>

5 0

3 years ago

Que alguien me ayude cuanto hace q yo sali de escuela

swat32

Answer:

a=20 m²

b=21 cm²

c=36 dm²

Step-by-step explanation:

A) 12²+16²=a²

144+256=a²

400=a²

√400=a

a=20

B) 28²+b²=35²

784+b²=1225

b²=1225-784

b²=441

b=√441

b=21

C) c²+15²=39²

c²+225=1521

c²=1296

c=√1296

c=36

7 0

3 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

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 $p_2$ .