Question Which numbers are equivalent to 25,000−1.56×103? Select all correct numbers.

Express 4x^2 − 12x in the form (2x + a)^2 + b.

gladu [14]

Answer: $(2x-3)^2-9\\$

a=-3

b=-9

Step-by-step explanation:

$4x^2-12x=(2x)^2-2*2x*3+9-9\\=(2x-3)^2-9\\$

8 0

3 years ago

Sheila is baking a few cakes for the bake sale for her school. Each cake requires 2 ½ cups of sugar. How many cakes can she bake

Ber [7]

Answer:

The answer is 2 cakes.

Step-by-step explanation:

What we know:

-Each cake requires 2 1/2 cups of sugar.

-Sheila has 7 1/3 cups of sugar.

What we need to know:

-How many cakes can Sheila make with 7 1/3 cups of sugar.

Remember:

-You must make a whole cake.

Show your work:

-I am going to convert the fraction into a decimal.

-2 1/2=2.5 and 7 1/3=7.3

-2 cakes will require 5 cups of sugar and you will remain with 2.3 cups of sugar.

-You cannot use 2.3 cups of sugar to make a cake. It must be in full.

5 0

2 years ago

Julie’s mom paid $2.70 for 5 pounds of bananas at Supermart. Mario’s mom paid $2.60 for 4 pounds of Bananas at Market One. Write

serg [7]

Answer: (Supermart) Equation 1 = $2.70÷5 lb = $0.54 per lb

(Market One) Equation 2 = $2.60÷4 lb = $0.65 per lb

Since the unit price of bananas per lb in Supermart is less than the unit price of bananas per lb in Market One, Supermart has the better price.

7 0

2 years ago

Read 2 more answers

PLZZ HELP 10 points thank you

Paraphin [41]

B < 11 = the board is shorter than 11 cm.
B > 11 = the box contains more then 11 books.
B < 12 = there are fewer than 12 beetles in a jar
B > 12 = the building is taller than 12 ft.

I hope this helped :)

4 0

3 years ago

One thousand independent rolls of a fair die will be made. Compute an approximation to the probability that the number 6 will ap

pashok25 [27]

Answer:

0.19

Step-by-step explanation:

We are given that

Number of rolls of die=n=1000

Let the event of six coming up be success.Then, in each trial , the probability of success =p=P(success)=P(6)= $\frac{1}{6}$

Let X be the random variable for the number of sixes in the 1000 rolls of die.

Then, $X\sim Binom(1000,\frac{1}{6})$

Since, n is very large,the binomial random variable can be approximated as normal random variable.

Mean, $\mu=np=1000\times \frac{1}{6}=166.67$

Variance= $\sigma^2=np(1-p)=1000\times \farc{1}{6}\times (1-\frac{1}{6})=1000\times \frac{5}{36}=138.89$

$X\sim N(166.67,138.89)$

$P(150\leq X\leq 200)=P[\frac{150-166.67}{11.79}\leq \frac{X-\mu}{\sigma}\leq \frac{200-166.67}{11.79}]$

= $P[-1.41\leq Z\leq 2.83]=P[Z\leq 2.83]-P[Z$

= $\phi(2.83)-\phi(-1.41)$

=0.9977-0.0793=0.9184

Thus, the probability that the number 6 appears between 150 to 200 times=0.92

Now, given that 6 appears exactly 200 times .

Therefore, other number appear in other 800 rolls .

We have to find the probability that the number 5 will appear less than 150 times.

Therefore, for 800 rolls, let the event of 5 coming up be success.

Then , p=P(success)=P(5)= $\frac{1}{5}$

Let Y be the random variable denoting the number of times 5 coming up in 800 rolls.

Then, $Y\sim bin(800,\frac{1}{5})$

Mean, $\mu=np=800\times \frac{1}{5}=160$

Variance, $\sigma^2=np(1-p)=800\times \frac{1}{5}(1-\frac{1}{5})=128$

$Y\sim N(160,128)$ because n is large

$P(Y$

$P(Y$

Hence, the probability that the number 5 will appear less than 150 times given that 6 appeared exactly 200 times=0.19

4 0

3 years ago