Solve for x: 1 < x + 3 < 4 please help

Each morning papa noted the birds feeding on his birdfedder . So far this month he had seen 59 blue jays ,68 black crows , 12 re

Oduvanchick [21]

To solve this question, you need to first look at the total amount of birds and then at the percentage of the blue jays that fed on the bird feeder out of all the other birds.

To calculate the total you add them all together:
59+68+12+1 = 140 birds

Then the percentages of the blue jays that fed on his bird feeder are:
59/140 *100 = 42.143%

You want to see how many blue jays we can expect to see out of 300 birds.
So you divide 300 by 100 to see what 1% is = 3 birds

Then you multiply this amount (1% = 3 birds) by 42.134% which is the percentage of blue jays we can expect to see out of the 300 birds.
3*42.134 = 126.43

You round this off to 126 because you cant expect to see half a bird. Even if the answer had been 126.9, you would have rounded it of to 126 because no matter how close it is to the next whole number; you either have a whole bird or you don't.

Final Answer = 126

5 0

3 years ago

Ice on a lake gets thicker as long as the temperature is below freezing. On a certain lake it took 8 days for the ice to reach a

spayn [35]

The average rate of growth of the ice can be found by dividing the thickness by the number of days of growth

The growth rate of the ice over the 8 days is 0.375 inches/day

Reason:

The known information on the ice are;

The time it took the ice to reach a thickness of 3 inches = 8 days

Required:

The (average) rate that the ice is growing in c=inches per day

Solution:

$Rate \ of \ growth \ of \ ice= \dfrac{Increase \ in \ thickness}{ Time} = \dfrac{\Delta \ Thickness}{\Delta T}$

Therefore;

$Rate \ of \ growth \ of \ ice= \dfrac{3 \ inches}{ 8 \ days} = 0.375 \ inches/day$

The (average) rate of growth of the ice over the 8 days is 0.375 inches per day

Learn more here:

brainly.com/question/17255666

4 0

2 years ago

Two urns both contain green balls and red balls. Urn I contains 6 green balls and 4 red balls and Urn II contains 8 green balls

34kurt

Presumably, whatever is drawn from Urn I is independent of what is drawn from Urn II. This means

$P(\text{red from Urn 1 AND red from Urn 2})=P(\text{red from Urn 1})\cdot P(\text{red from Urn 2})$

We have

$P(\text{red from Urn 1})=\dfrac4{10}=\dfrac25$

$P(\text{red from Urn 1})=\dfrac7{15}$

so the probability of drawing red balls from both urns is $\dfrac25\cdot\dfrac7{15}=\dfrac{14}{75}$

3 0

3 years ago

Mei runs 92.16 miles in 3 weeks. If she runs 6 days each week, what is the average distance she runs each day? THIS IS DUE IN 10

MissTica

Step-by-step explanation:

Mei runs 6 days each week.

=> Mei runs 18 days in 3 weeks.

Since she ran 92.16 miles in 18 days,

She ran 92.16 miles / 18 days

= 5.12 miles/day.

6 0

3 years ago

Sarah is doing research on the average age of first-year resident physicians. She wants her estimate to be accurate to within 1

WINSTONCH [101]

Answer: The number of first-year residents she must survey to be 95% confident= 263

Step-by-step explanation:

When population standard deviation ( $\sigma$ ) is known and margin of error(E) is given, then the minimum sample size (n) is given by :-

$n=(\dfrac{z^*\sigma}{E})^2$ , z* = Two-tailed critical value for the given confidence interval.

For 95% confidence level , z* = 1.96

As, $\sigma$ = 8.265, E = 1

So, $n= (\dfrac{1.96\times8.265}{1})^2 =(16.1994)^2\\\\= 262.42056036\approx263\ \ \ [\text{Rounded to the next integer}]$

Hence, the number of first-year residents she must survey to be 95% confident= 263

7 0

3 years ago