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3 years ago

How would you solve this equation when -14-7=-7?

Mathematics

1 answer:

uysha [10]3 years ago

8 0

Answer:

r = -8/5

Step-by-step explanation:

-7 log4 (-10r) = -14

Divide each side by -7

-7/-7 log4 (-10r) = -14/-7

log4 (-10r) = 2

Raise each side to the base of 4

4^ log4 (-10r) = 4^2

-10r = 16

Divide each side by -10

-10r/-10 = 16/-10

r = -116/10

r = -8/5

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5. You are a packer for Acme Bottling Company. You fill cartons with bottles,

rjkz [21]

Answer:

$459.65

Step-by-step explanation:

The total number of cartons you filled was ...

210 +175 +224 +160 +216 +90 = 1075

so your pay for filling those is ...

$0.41 × 1075 = $440.75

You get an additional amount of pay for the cartons filled on Saturday:

$0.21 × 90 = $18.90

So your pay for the week is ...

$440.75 +18.90 = $459.65

6 0

4 years ago

Read 2 more answers

suppose a rabbit population of 10 rabbits trippled every 2 months. write an exponential function that represents the population

tangare [24]

Answer:

exponential function that represents the population of rabbits : $10*3^{\frac{x}{2} }$

number of rabbits after 2 years : $10*3^{12}$

Step-by-step explanation:

initially there are 10 rabbits. since the will be trippled every 2 months, i have multiplied it with $3^{\frac{x}{2} }$ . I have divided x by 2 because, here x represents number of months and it will be trippled only AFTER EVERY 2 MONTHS.

variable x: it represents number of months.

two years= 2*12 months

now substitute x= 2*12 in the above function.

number of rabbits after 2 years = $10*3^{\frac{2*12}{2} } = 10*3^{12}$

7 0

4 years ago

A local bank needs information concerning the savings account balances of its customers. A random sample of 15 accounts was chec

PSYCHO15rus [73]

Answer:

(513.14, 860.36) is a 98% confidence interval for the true mean

Step-by-step explanation:

We have a small sample size of n = 15, the mean balance was $\bar{x} = 686.75$ with a standard deviation of s = 256.20. The confidence interval is given by $\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}})$ where $t_{\alpha/2}$ is the 100 $(\alpha/2)$ th quantile of the t distribution with n-1=15-1=14 degrees of freedom. As we want a $100(1-\alpha)$ % = 98% confidence interval, we have that $\alpha = 0.02$ and the confidence interval is $686.75\pm t_{0.01}(\frac{256.20}{\sqrt{15}})$ where $t_{0.01}$ is the 1st quantile of the t distribution with 14 df, i.e., $t_{0.01} = -2.6245$ . Then, we have $686.75\pm (-2.6245)(\frac{256.20}{\sqrt{15}})$ and the 98% confidence interval is given by (513.1379, 860.3621)