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3 years ago

What is the product? -4 x (8 -1 -5 9)

Mathematics

2 answers:

SOVA2 [1]3 years ago

5 0

The product is 152x :))

snow_lady [41]3 years ago

4 0

Step-by-step explanation:

If you mean that - 4×(8-1-5×9)=152

Or - 4×(8×-1×-5×9)=-1440

Or - 4×(-8+-1+-5+9)=-44

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Okay, help me out here!!

luda_lava [24]

Answer:

65.8

Step-by-step explanation:

(2.6*4.7)/2=6.11

11*4.7=51.7

(4.7(17-(2.6+11)))/2=7.99

6.11+51.7+7.99=65.8

3 0

4 years ago

700=132.69x-25.96 <br> solve please

Damm [24]

Answer:

x = 5.5 (rounded)

Step-by-step explanation:

Equation: 700 = 132.69x - 25.96

Add 25.96 to both sides: 700+25.96 = 132.69x -25.96 + 25.96

Simplify: 725.96 = 132.69x

Isolate x

Divided both sides by 132.69: $\frac{725.96}{132.69} = \frac{132.69x}{132.69}$

Simplify: x = 5.5 (rounded)

7 0

3 years ago

Factor the expression 2x^3-10x^2-5x+25. Show your work.

azamat

Step 1. Factor out common terms in the first two terms, then in the last two terms.

2x^2(x - 5) -5(x - 5)

Step 2. Factor out the common term x - 5

(x - 5)(2x^2 - 5)

4 0

3 years ago

Can someone help me with questions 2 and 3 pls?

bija089 [108]

1 cup = 1/2 pint, so where it says 4 cups we'll just read 2 pints.

1. In one show we serve 30 × 2 pints of small and 42 × 6 pints of regular.

Total = 30 × 2 + 42 × 6 = 60 + 252 = 312 pints

Answer: 312 pints

2. 8 pints per gallon, 2 gallons per container makes 16 pints per container.

312 pints / 16 pints per container = 19.5 containers

We round up since we need a whole number of containers.

Answer: 20 containers

We can't really see question 3.

7 0

3 years ago

A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and

pav-90 [236]

Answer:

ans=13.59%

Step-by-step explanation:

The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean $\mu$ and standard deviation $\sigma$ , we have these following probabilities

$Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827$

$Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545$

$Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973$

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So $\mu = 53, \sigma = 11$

So:

$Pr(53-11 \leq X \leq 53+11) = 0.6827$

$Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545$

$Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973$

-----------

$Pr(42 \leq X \leq 64) = 0.6827$

$Pr(31 \leq X \leq 75) = 0.9545$

$Pr(20 \leq X \leq 86) = 0.9973$

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have $Pr(31 \leq X \leq 75) = 0.9545 = 0.9545$ subtracted by $Pr(42 \leq X \leq 64) = 0.6827$ is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.

To find just the percentage above the mean, we divide this value by 2

So:

$P = {0.9545 - 0.6827}{2} = 0.1359$

The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.

4 0

4 years ago