PLEASE HELP

Alex dumped out his piggy bank and there were only

maria [59]

Answer:

12 nickels

36 dimes

52 quarters

Step-by-step explanation:

Let x = no. of nickels

3x = no. of dimes

x + 40 = no. of quarters

5x = value of the nickels since each nickel is 5 cents

10(3x) = 30x = value of the dimes since each dime is 10 cents

25(x + 40) = 25x + 1000 = value of the quarters since each quarter is 25 cents

Now, 5x + 30x + 25x + 1000 = 1720 ($17.20 = 1720 cents)

60x + 1000 = 1720

60x = 720

x = 12 nickels

3x = 36 dimes

x + 40 = 52 quarters

Check: 12 nickels are worth $.60

36 dimes are worth $3.60

52 quarters are worth $13

.60 + 3.60 + 13 = 17.20

4 0

3 years ago

Which of the following are examples of exponential growth?

____ [38]

2^x( not in math type) (so your answer would be 2 to the x power and x is whatever number you out in)

6 0

3 years ago

Hey can you please help me posted picture of question

dangina [55]

If you observe the graphs, you can see that the graph of G(x) is obtained by moving F(x) 1 unit to left.

Moving one unit to left is equivalent to adding 1 to x in the base of the expression in given case.

So,

G(x) = F(x+1)

G(x) = (x+1)²

Thus the correct answer is option D

8 0

4 years ago

9 less than 5 times some number

valina [46]

nine less than five times a number is 26

6 0

3 years ago

Read 2 more answers

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

(c) 90.53%

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .