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3 years ago

15

Find each sum simplify if possible

1 answer:

notsponge [240]3 years ago

7 0

That's simplied already, unless they want a decimal then it's .6 or something

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Maya wants to buy a house for $275,000 by taking out a 30-year fixed-rate mortgage with an interest rate of 6%. She plans on mak

Marina86 [1]

Hello,

First you have to multiply $275,000 x 30 x 6% = $495,000.

then 495,000 x 6% = 29,700.

That is the answer it is between 25,000 - 45,000....

Hope I Helped!!!

6 0

3 years ago

Read 2 more answers

Kenneth has a goal to work 287 hours each month at the pizza shop. So far he has worked 124

murzikaleks [220]

Answer:

163

Step-by-step explanation:

6 0

2 years ago

40% of Oatypop cereal boxes contain a prize. Hannah plans to keep buying cereal until she gets a prize. What is the probability

aleksley [76]

Answer:

$(0.6)^{n}+n(0.4)(0.6)^{n-1}+\frac{n(n-1)(0.4)^{2}0.6^{n-2}}{2} +\frac{n(n-1)(n-2)0.4^{3}0.6^{n-3}}{6}$

Step-by-step explanation:

We will use the binomial distribution. Let X be the random variable representing the no. of boxes Hannah buys before betting a prize.

Our success is winning the prize, p =40/100 = 0.4

Then failure q = 1-0.4 = 0.6

Hannah keeps buying cereal boxes until she gets a prize. Then n be no. times she buys the boxes.

P(X ≤ 3) = P(X=0) +P(X=1)+P(X=2)+P(X=3)

= $\binom{n}{0}p^{0}q^{n-0}$ + $\binom{n}{1}p^{1}q^{n-1}$ + $\binom{n}{2}p^{2}q^{n-2}$ + $\binom{n}{3}p^{3}q^{n-3}$

= $q^{n}+npq^{n-1}$ + $\frac{n(n-1)(p)^{2}q^{n-2}}{2}$ + $\frac{n(n-1)(n-2)p^{3}q^{n-3}}{6}$

= $(0.6)^{n}+n(0.4)(0.6)^{n-1}+[tex]\frac{n(n-1)(0.4)^{2}0.6^{n-2}}{2} +\frac{n(n-1)(n-2)0.4^{3}0.6^{n-3}}{6}$

7 0

3 years ago

Read 2 more answers

Try solving this: (5 3/4) (2)

denpristay [2]

Answer:

11 1/2

Step-by-step explanation:

7 0

3 years ago

Find values of  that satisfy the equation for 0º    360º and 0  θ  2 . Give answers in degrees and radians.

suter [353]

Answer:

$\theta = 30^\circ, 330^\circ$

$\theta = \frac{\pi}{6}, \frac{11\pi}{6}$

Step-by-step explanation:

Given [Missing from the question]

Equation:

$cos\theta = \frac{\sqrt 3}{2}$

Interval:

$0 \le \theta \le 360$

$0 \le \theta \le 2\pi$

Required

Determine the values of $\theta$

The given expression:

$cos\theta = \frac{\sqrt 3}{2}$

... shows that the value of $\theta$ is positive

The cosine of an angle has positive values in the first and the fourth quadrants.

So, we have:

$cos\theta = \frac{\sqrt 3}{2}$

Take arccos of both sides

$\theta = cos^{-1}(\frac{\sqrt 3}{2})$

$\theta = 30$ --- In the first quadrant

In the fourth quadrant, the value is:

$\theta = 360 -30$

$\theta = 330$

So, the values of $\theta$ in degrees are:

$\theta = 30^\circ, 330^\circ$

Convert to radians (Multiply both angles by $\pi/180$ )

So, we have:

$\theta = \frac{30 * \pi}{180}, \frac{330 * \pi}{180}$

$\theta = \frac{\pi}{6}, \frac{33 * \pi}{18}$

$\theta = \frac{\pi}{6}, \frac{11 * \pi}{6}$

$\theta = \frac{\pi}{6}, \frac{11\pi}{6}$

8 0

2 years ago