5x5x5x5x5x5x5x5x10x10x00.0001

The previous balance in your check register is $234. 67. You make a deposit of $150. 0. You write a check for $63. 23. What is y

Leno4ka [110]

Answer:

321.44

Step-by-step explanation:

(234.67+150)-63.23

8 0

2 years ago

A manufacturer of shipping boxes has a box shaped like a cube. The side length is (5a + 4b). What is the volume of the box in te

rodikova [14]

Volume of a square = (s)(s)(s) and s=5a+4b Therefore: (5a+4b)(5a+4b)(5a+4b)= (25a^2+40ab+16b^2)(5a+4b)= (125a^3+200a^2b+80ab^2+100a^2b+160ab^2+64b^3)= (125a^3+300a^2b+240ab^2+64b^3) You just multiple the first time the second and then do so again with the combination of the first two times the second. A little cleaning up and you are left with an equation in terms of a and b.

7 0

3 years ago

Find the value of the ratio. Can someone help please???

USPshnik [31]

Answer:

SOH CAH TOA.....Cosine=adj/hyp...... __cosX=16/30___

SOH CAH TOA,,,,,, Tangent= opp/adj-------- Tan X=36/27

Step-by-step explanation:

3 0

2 years ago

An equation _____ has one solution. never sometimes always

valentina_108 [34]

Answer:

the answer to that is sometimes

7 0

3 years ago

A normally distributed population has mean 57,800 and standard deviation 750. Find the probability that a single randomly select

Stels [109]

Answer:

(a) Probability that a single randomly selected element X of the population is between 57,000 and 58,000 = 0.46411

(b) Probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = 0.99621

Step-by-step explanation:

We are given that a normally distributed population has mean 57,800 and standard deviation 75, i.e.; $\mu$ = 57,800 and $\sigma$ = 750.

Let X = randomly selected element of the population

The z probability is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

(a) So, P(57,000 <= X <= 58,000) = P(X <= 58,000) - P(X < 57,000)

P(X <= 58,000) = P( $\frac{X-\mu}{\sigma}$ <= $\frac{58000-57800}{750}$ ) = P(Z <= 0.27) = 0.60642

P(X < 57000) = P( $\frac{X-\mu}{\sigma}$ < $\frac{57000-57800}{750}$ ) = P(Z < -1.07) = 1 - P(Z <= 1.07)

= 1 - 0.85769 = 0.14231

Therefore, P(31 < X < 40) = 0.60642 - 0.14231 = 0.46411 .

(b) Now, we are given sample of size, n = 100

So, Mean of X, X bar = 57,800 same as before

But standard deviation of X, s = $\frac{\sigma}{\sqrt{n} }$ = $\frac{750}{\sqrt{100} }$ = 75

The z probability is given by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

Now, probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = P(57,000 < X bar < 58,000)

P(57,000 <= X bar <= 58,000) = P(X bar <= 58,000) - P(X bar < 57,000)

P(X bar <= 58,000) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ <= $\frac{58000-57800}{\frac{750}{\sqrt{100} } }$ ) = P(Z <= 2.67) = 0.99621

P(X < 57000) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{57000-57800}{\frac{750}{\sqrt{100} } }$ ) = P(Z < -10.67) = P(Z > 10.67)

This probability is that much small that it is very close to 0

Therefore, P(57,000 < X bar < 58,000) = 0.99621 - 0 = 0.99621 .

7 0

3 years ago