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2 years ago

Help pleaseeeeeeeeeee

Mathematics

1 answer:

vova2212 [387]2 years ago

6 0

Answer:

B & C

Step-by-step explanation:

The two binomials are equal to zero to find the roots.

Hope this helps

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Will give crown will post more

marta [7]

Answer:

2, 7, 27, 42

Step-by-step explanation:

y = 5*1 -3 =2

y = 5*2 -3 =7

y = 5*6 -3 = 27

y = 5*9 -3 = 42

4 0

2 years ago

Read 2 more answers

On friday, 2/5 of the students wore blue shirts and 5/12 of the students wore white shirts. What fraction of the student wore ei

SIZIF [17.4K]

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3 years ago

PLEASE HELP ME WITH QUESTION #7 !!!! PLEASE!!!!! 20 POINTS!!!

pashok25 [27]

let x = total charge

Jane = X=5 + 0.10m

Joe = x=4+0.20m

5 0

3 years ago

the patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.9 days and a standard de

NNADVOKAT [17]

The 85th percentile is the cutoff time $t$ such that

$\mathbb P(X$

In other words, the 85th percentile refers to the time needed to belong to the top 15% of the distribution; more generally, the $n$ percentile is the top $(100-n)\%$ of the distribution.

Anyway, to find this value of $t$ , transform $X$ to a random variable $Z$ with the standard normal distribution using

$Z=\dfrac{X-\mu}\sigma$

where $\mu$ is the mean of $X$ and $\sigma$ is the standard deviation of $X$ .

$\mathbb P(X$

Here $t^*$ is used to denote the z-score corresponding to the cutoff time $t$ . Referring to a z-score table, you find that this occurs for $t^*\approx1.036$ . So,

$\dfrac{t-5.9}{2.1}=t^*\implies t=5.9+2.1t^*\approx8.076$

7 0

3 years ago

Suppose that the warranty cost of defective widgets is such that their proportion should not exceed 5% for the production to be

Vedmedyk [2.9K]

Answer: 63 defective widgets

Step-by-step explanation:

Given that the proportion should not exceed 5%, that is:

p< or = 5%.

So we take p = 5% = 0.05

q = 1 - 0.05 = 0.095

Where q is the proportion of non-defective

We need to calculate the standard error (standard deviation)

S = √pq/n

Where n = 1024

S = √(0.05 × 0.095)/1024

S = 0.00681

Since production is to maximize profit(profitable), we need to minimize the number of defective items. So we find the limit of defective product to make this possible using the Upper Class Limit.

UCL = p + Za/2(n-1) × S

Where a is alpha of confidence interval = 100 -90 = 10%

a/2 = 5% = 0.05

UCL = p + Z (0.05) × 0.00681

Z(0.05) is read on the t-distribution table at (n-1) degree of freedom, which is at infinity since 1023 = n-1 is large.

Z a/2 = 1.64

UCL = 0.05 + 1.64 × 0.00681

UCL = 0.0612

Since the UCL in this case is a measure of proportion of defective widgets

Maximum defective widgets = 0.0612 ×1024 = 63

Alternatively

UCL = p + 3√pq/n

= 0.05 + 3(0.00681)

= 0.05 + 0.02043 = 0.07043

UCL =0.07043

Max. Number of widgets = 0.07043 × 1024

= 72

7 0

3 years ago