Someone please help me?!?!

Insert 2 sets of parentheses to make each sentence true: 2 x 14 - 9 - 17 - 14 = 7

Mandarinka [93]

It’s most definitely C

6 0

2 years ago

How do I do this please help me?

TEA [102]

4. First find how much sauce is used for 1 person:

2 ÷ 7 = 0.285714285714286

0.285714285714286cups of sauce will make spaghetti for one person.

0.285714285714286 • 210 = 60

60 cups of sauce for 210 people.

5. Found out how much he read in one minute:

40 ÷ 50 = 0.8

0.8 • 80 = 64

In 80 minutes, she will read 64 pages.

6. Found out how much he earns after one month:

9 ÷ 6 = 1.5

1.5 • 12 = 18

18 • 2 = 36

After two years, he will have 36 days of vacationing time

7 0

2 years ago

a dish has 21,245,673 bacteria in it. The population of bacteria will decline by 80% every day. How many bacteria will be presen

Ivahew [28]

Answer:

33993.0768

Step-by-step explanation:

You can multiply the number of bacteria by (0.2^4). The 0.2 is for the 20% that left after each day. The (^4) is for the 4 days that have passed.

6 0

2 years ago

Math problem can somebody help!!!!

abruzzese [7]

Answer:

if am correct your answer will be letter d

Step-by-step explanation:

6 0

1 year ago

] It is claimed that 42% of US college graduates had a mentor in college. For a sample of college graduates in Colorado, it was

Bad White [126]

Answer:

$z=\frac{0.480 -0.42}{\sqrt{\frac{0.42(1-0.42)}{1045}}}=3.930$

The p value for this case would be given by:

$p_v =P(z>3.930)=0.0000443$

For this case the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis so then we can conclude that the true proportion is significantly higher than 0.42

Step-by-step explanation:

Information given

n=1045 represent the random sample selected

X=502 represent the college graduates with a mentor

$\hat p=\frac{502}{1045}=0.480$ estimated proportion of college graduates with a mentor

$p_o=0.42$ is the value that we want to test

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true proportion is higher than 0.42, the system of hypothesis are.:

Null hypothesis: $p \leq 0.42$

Alternative hypothesis: $p > 0.42$

The statistic is given by:

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

Replacing the info we got:

$z=\frac{0.480 -0.42}{\sqrt{\frac{0.42(1-0.42)}{1045}}}=3.930$

The p value for this case would be given by:

$p_v =P(z>3.930)=0.0000443$

For this case the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis so then we can conclude that the true proportion is significantly higher than 0.42

7 0

2 years ago