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3 years ago

What is the GCF of 14 and 35

1 answer:

5 0

Factor each number:
14=2*7
35=5*7

7 is a common factor, and no other (except 1), so 7 is the greatest common factor (GCF) of 14 and 35.

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WILL GIVE BRAINLIEST! Simplify 350−−√+332−−√ Keep in simplified radical form.

Bess [88]

Answer:

350

Step-by-step explanation:

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3 years ago

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3 years ago

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seropon [69]

Given:

32 and 28 are the two numbers.

To find:

(a) The GCF of 32 and 28.

(b) Using the GCF, find the factor of (32-28).

Solution:

(a)

The factor forms of the given numbers are:

$32=2\times 2\times 2\times 2\times 2$

$28=2\times 2\times 7$

It is clear that, 2 and 2 are the common factors in both.

$GCF(32,28)=2\times 2$

$GCF(32,28)=4$

Hence, the GCF of 32 and 28 is 4.

(b)

Using the GCF we need to find the factor of (32-28).

$32-28=4\times 8-4\times 7$

Taking out the GCF, we get

$32-28=4\times (8-7)$

Hence, $32-28=4\times (8-7)$ .

8 0

3 years ago

I will mark the best answer as brainliest.

dimulka [17.4K]

Answer: They didn't solve the problem all the way, you need to find the square root of 674. The final answer is 25.96

Step-by-step explanation:

$a^{2}$ + $b^{2}$ = $c^{2}$

$7^{2}$ + $25^{2}$ = 674

$\sqrt{674}$ = 25.96

6 0

3 years ago

Read 2 more answers

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