Pls answer :). Plsssss

Sara pays $4.50 for 3 pounds of bananas and 2 pounds of oranges. One pound of oranges costs $0.75 more than one pound of bananas

Svetlanka [38]

The price of one pound of banana = b

As one pound of oranges costs $0.75 more than one pound of bananas, so, the price of one pound of orange = b+0.75.

The price of 3 pounds of banana = 3b,

and the price of 2 pounds of oranges = 2(b+0.75).

Now, as she pays $4.50 for 3 pounds of bananas and 2 pounds of oranges.

So, for this situation the required equation is

3b + 2(b+0.75) =4.5

On solving this equation, we have

$\Rightarrow 3b+2b+2\times0.75=4.5$

$\Rightarrow 5b +1.5=4.5$

$\Rightarrow 5b =4.5-1.5=3$

$\Rightarrow b =\frac 3 5=0.6$

Hence, the price of one pound of banana = $ 0.60

and the price of one pound of orange

=b+0.75= 0.60+0.75=$1.35.

8 0

4 years ago

Which point on the graph represents the y-intercept? On a coordinate plane, a line goes through points V (negative 4, 4), W (0,

scoundrel [369]

Answer:

The y-intercept is the point where x equals 0, so point W(0,2) would be the y-intercept.

4 0

4 years ago

Read 2 more answers

Evaluate each expression.

Virty [35]

Answer:

720

12

120

Step-by-step explanation:

Factorials

We must recall that the factorial of a number n (positive or zero) is the product of all the integers from n down to 1

$n!= n(n-1)(n-2)...1$

Let's evaluate the given expressions

$6!=6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1=720$

Similarly

$3!\cdot 2!=(3\cdot 2\cdot 1)\cdot (2\cdot 1)=6\cdot 2=12$

Finally

$\displaystyle \frac{6!}{3!}=\frac{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{3\cdot 2\cdot 1}$

$\displaystyle \frac{6!}{3!}= \frac{720}{6}=120$

3 0

4 years ago

Read 2 more answers

23. Tara is driving 30 miles per hour. How long will it take her to travel 60 miles?

lys-0071 [83]

Answer:

2 hours

Step-by-step explanation:

30×1=30

30×2=60

4 0

3 years ago

A data set lists earthquake depths. The summary statistics are

irinina [24]

Answer:

Step-by-step explanation:

The summary of the given statistics data include:

sample size n = 400

sample mean $\overline x$ = 6.86

standard deviation = 4.37

Level of significance ∝ = 0.01

Population Mean $\mu$ = 6.00

Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.

To start with the hypothesis;

The null and the alternative hypothesis can be computed as :

$H_o: \mu = 6.00 \\ \\ H_1 : \mu \neq 6.00$

The test statistics for this two tailed test can be computed as:

$z= \dfrac{\overline x - \mu}{\dfrac{\sigma}{\sqrt {n}}}$

$z= \dfrac{6.86 - 6.00}{\dfrac{4.37}{\sqrt {400}}}$

$z= \dfrac{0.86}{\dfrac{4.37}{20}}$

z = 3.936

degree of freedom = n - 1

degree of freedom = 400 - 1

degree of freedom = 399

At the level of significance ∝ = 0.01

P -value = 2 × (z < 3.936) since it is a two tailed test

P -value = 2 × ( 1 - P(z ≤ 3.936)

P -value = 2 × ( 1 -0.9999)

P -value = 2 × ( 0.0001)

P -value = 0.0002

Since the P-value is less than level of significance , we reject $H_o$ at level of significance 0.01

Conclusion: There is sufficient evidence to conclude that the original claim that the mean of the population of earthquake depths is 5.00 km.

7 0

3 years ago