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3 years ago

X- 14x + 40

Mathematics

2 answers:

ivanzaharov [21]3 years ago

6 0

Answer: b. (x−4)(x−10)

Step-by-step explanation:

Let's factor x^2−14x+40

x^2−14x+40

The middle number is -14 and the last number is 40.

Factoring means we want something like

(x+_)(x+_)

Which numbers go in the blanks?

We need two numbers that...

Add together to get -14

Multiply together to get 40

Can you think of the two numbers?

Try -4 and -10:

-4+-10 = -14

-4*-10 = 40

Fill in the blanks in

(x+_)(x+_)

with -4 and -10 to get...

(x-4)(x-10)

Answer:

(x−4)(x−10)

Vinil7 [7]3 years ago

5 0

The answer is : B. (X-10)(x-4)

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Whole numbers greater than 10 but less than 20

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2 years ago

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A. What would be the cost of the Standard Daily Rate plus Mileage plan?

Jobisdone [24]

Answer:

Part a) $\$190$

Part b) $\$315$

Part c) Standard Daily Rate plus Mileage plan

Step-by-step explanation:

Part a) What would be the cost of the Standard Daily Rate plus Mileage plan?

Let

y ----> the total cost

x ----> the number of kilometers

we have that

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$y=70+0.30x$

For x=400 km

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$y=70+0.30(400)$

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Part b) What would be the cost of the Unlimited Mileage plan?

we know that

The Unlimited Mileage plan for a Luxury car is $105 per day (see the table)

The trip are three days

To find the total cost multiply $105 by 3

$3(\$105)=\$315$

Part c) Which is the better plan?

Compare

$\$190 < \$315$

therefore

For this trip the better plan is the Standard Daily Rate plus Mileage plan

7 0

3 years ago

What is 2.50x + 4x = 39

34kurt

2.5x + 4x = 39
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3 years ago

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Use the standard norml distribution or the t-distribution to construct a 99% confidence interval for the population mean. Justif

dalvyx [7]

Answer:

E) we will use t- distribution because is un-known,n<30

the confidence interval is (0.0338,0.0392)

Step-by-step explanation:

Step:-1

Given sample size is n = 23<30 mortgage institutions

The mean interest rate 'x' = 0.0365

The standard deviation 'S' = 0.0046

the degree of freedom = n-1 = 23-1=22

99% of confidence intervals $t_{0.01} =2.82$ (from tabulated value).

$The mean value = 0.0365$

$x±t_{0.01} \frac{S}{\sqrt{n-1} }$

$0.0365±2.82 \frac{0.0046}{\sqrt{23-1} }$

$0.0365±2.82 \frac{0.0046}{\sqrt{22} }$

$0.0365±2.82 \frac{0.0046}{4.690 }$

using calculator

$0.0365±0.00276$

Confidence interval is

$(0.0365-0.00276,0.0365+0.00276)$

$(0.0338,0.0392)$

the mean value is lies between in this confidence interval

(0.0338,0.0392).

Answer:-

using t- distribution because is unknown,n<30,and the interest rates are not normally distributed.

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2 years ago

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