Help please 5 points

Please please please tell the answer

OlgaM077 [116]

Answer:

a. 0

b. 4^1

c. 14

d. 8

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

A store had 50 bottles of olive oil. Each week, 40% of the olive oil bottles were sold and 20 new bottles arrived in shipments.

Wittaler [7]

Answer:

f(n) = f(n − 1) ⋅ 0.6 + 20, n > 0

Step-by-step explanation:

Each week 40% of the olive oil bottles were sold. This means, each week 60% of the bottles were left. In addition to these, 20 new bottles arrived in shipments.

So, every week the number of shipments was:

60% of the shipments of previous week + 20

In equation form this can written as:

Shipments in a week = 60% of shipments of previous week + 20

For, nth week we can re-write this equation as:

f(n) = 60% of f(n - 1) + 20

or

f(n) = f(n - 1) ⋅ 0.60 + 20

For example for first week, replace n by 1 to find the number of shipments in week 1 and same goes for next weeks.

8 0

2 years ago

Read 2 more answers

How many days will it take for a culture of bacteria to increase from 200 to 50000 if the growth rate per day is 93.2%

densk [106]

Answer:

89 days

Step-by-step explanation:

5 0

2 years ago

A box of candy is shaped like a triangular pyramid the volume of the candy box measures 120 in squared 3 the area of the base of

Zolol [24]

Answeh:

Height

A: Base Area

Volume = Ah/3

base=18

volume area of the box =120

=720

Step-by-step explanation:

8 0

2 years ago

A random sample of 144 recent donations at a certain blood bank reveals that 81 were type A blood. Does this suggest that the ac

Reptile [31]

Answer:

Null hypothesis: $p=0.4$

Alternative hypothesis: $p \neq 0.4$

$z=\frac{0.5625 -0.4}{\sqrt{\frac{0.4(1-0.4)}{144}}}=3.98$

$p_v =2*P(z>3.98)=0.0000689$

Since the p value is very low compared to the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true percent of people with type A of blood is significantly different from 0.4 or 40%

Step-by-step explanation:

Information given

n=144 represent the random sample taken

X=81 represent the number of people with type A blood

$\hat p=\frac{81}{144}=0.5625$ estimated proportion of people with type A blood

$p_o=0.4$ is the value that we want to verify

$\alpha=0.01$ represent the significance level

z would represent the statistic

$p_v{/tex} represent the p value Hypothesis to testWe want to test if the percentage of the population having type A blood is different from 40%.: Null hypothesis:[tex]p=0.4$

Alternative hypothesis: $p \neq 0.4$

the statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing the info given we got:

$z=\frac{0.5625 -0.4}{\sqrt{\frac{0.4(1-0.4)}{144}}}=3.98$

Now we can calculate the p value with this probability taking in count the alternative hypothesis:

$p_v =2*P(z>3.98)=0.0000689$

Since the p value is very low compared to the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true percent of people with type A of blood is significantly different from 0.4 or 40%

3 0

2 years ago