Tina wants to run 123 miles over

Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 60

Umnica [9.8K]

Sample mean : \overline{x}=10.6x=10.6

Standard deviation : s=1.7s=1.7

Significance level : \alpha:1-0.95=0.05α:1−0.95=0.05

Critical value : z_{\alpha/2}=1.96

Hence the 95% confidence interval for the number of chocolate chips per cookie for big chip cookies= (10.1989,\ 11.0011)(10.1989, 11.0011)

4 0

2 years ago

Brian has a gift card with a $250 balance. He buys several video games that cost $33 each. After the purchases, his gift card ba

Leno4ka [110]

Answer: x=6.6969

Step-by-step explanation:

250-(33x)=29

-33x=-221

x=6.6969

5 0

3 years ago

Read 2 more answers

Helpppppppppppppppppppppppppp

Virty [35]

Answer:

3:1

Step-by-step explanation:

There are 3 birch trees and one willow tree

4 0

3 years ago

The television station has received/A many complaints about/B

NARA [144]

The television station has received/A many complaints about/B the clothing advertisements, which some/C viewers condemn to be/D tasteless. No error/E

The correct option is D.

The television station has received many complaints about the clothing advertisements, which some viewers complained were tasteless.
Well, you could mean that Writing Question is a 'trick one' .

What is a television station called?

Most often the term "television station" refers to a station which broadcasts structured content to an audience or it refers to the organization that operates the station.
A terrestrial television transmission can occur via analog television signals or, more recently, via digital television signals.

Learn more about television station

brainly.com/question/2851169

#SPJ4

6 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

3 years ago