How wide is a rectangular strip of land with the length 3/4 mile and area 1/2 square mile?

Mary pays $24.00 for 5 pounds of apples. At the same price, how much would she pay

mina [271]

First calculate the price of one pound of apples
24/5
= $4.80
Then multiply it by 6
6 x 4.80
Your price is $28.80

4 0

3 years ago

Please help write the equation of the line.

AfilCa [17]

Answer:

x -2y -2=0

Step-by-step explanation:

using the formula, m= ( Y2 - Y1)/( X2-X1)

the gradient is found. the coordinates chosen are (2,0) and ( 0,-1)

the equation of the line is obtained using the formula

y -y1= m(x -x1)

3 0

2 years ago

Safiya deposited money into a bank account that earned 6.5% simple interest each year. After 312 years, she had earned $36.40 in

Svetllana [295]

Answer: If the time is 3 1/2 years (most reasonable assumption)

Her initial deposit was $160

Step-by-step explanation: The basic formula for simple interest is

<em>i = prt i </em>is interest, P is principal (deposit) r is the percentage (converted to a decimal) and t is time, the number of years.

For this question, you have to rearrange the equation to solve for p.

So divide both sides by rt and you have

p = i/rt

Substitute the values given

p = 36.40/(0.065)(3.5)

8 0

3 years ago

Round the following value to the nearest cent .<br>$506.106 <br>Im confused.

matrenka [14]

I would be 506.110 cause if it is over five then round it to ten

6 0

3 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