If A = lw, find the width of a rectangle with a length of 30 miles and an area of 210 square miles.

David will spend at most $39 on gifts. So far, he has spent $22. What are the possible additional amounts he will spend? Use c f

iren2701 [21]

Answer:

$c\leq \$17$

Step-by-step explanation:

Let

c -----> the additional amount (in dollars) David will spend

we know that

The word " at most" in this context means "less than or equal to"

The amount David has spent plus possible additional amounts he will spend must be less than or equal to $39

so

The inequality that represent this situation is

$22+c\leq 39$

solve for c

subtract 22 both sides

$c\leq 39-22$

$c\leq \$17$

The maximum amount he could spend is $17

7 0

2 years ago

if your car gets 26 miles per gallon how much does it cost to drive 420 miles when gas is $2.80 per gallon

Kitty [74]

Answer:

$45.22

Step-by-step explanation:

The number of miles per gallon can be represented by the ratio:

$\frac{miles}{gallon}=\frac{26}{1}$

Since we know the unit rate of 26 miles per gallon of gas, we can set up a proportion, or equivalent ratios, to find the number of gallons needed for 420 miles:

$\frac{26}{1}=\frac{420}{x}$ , where 'x' is the number of gallons of gas

cross-multiply and divide: 26x = 420 or x ≈ 16.15 gallons

If gas is $2.80 per gallon, take the total number of gallons needed and multiply by $2.80:

16.15 x $2.80 = $45.22

5 0

3 years ago

The mean SAT score in mathematics, M, is 600. The standard deviation of these scores is 48. A special preparation course claims

kupik [55]

Answer:

Step-by-step explanation:

The mean SAT score is $\mu=600$ , we are going to call it \mu since it's the "true" mean

The standard deviation (we are going to call it $\sigma$ ) is

$\sigma=48$

Next they draw a random sample of n=70 students, and they got a mean score (denoted by $\bar x$ ) of $\bar x=613$

The test then boils down to the question if the score of 613 obtained by the students in the sample is statistically bigger that the "true" mean of 600.

- So the Null Hypothesis $H_0:\bar x \geq \mu$

- The alternative would be then the opposite $H_0:\bar x < \mu$

The test statistic for this type of test takes the form

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}$

and this test statistic follows a normal distribution. This last part is quite important because it will tell us where to look for the critical value. The problem ask for a 0.05 significance level. Looking at the normal distribution table, the critical value that leaves .05% in the upper tail is 1.645.

With this we can then replace the values in the test statistic and compare it to the critical value of 1.645.

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}\\\\= \frac{| 600-613 |}{48/\sqrt(70}}\\\\= \frac{| 13 |}{48/8.367}\\\\= \frac{| 13 |}{5.737}\\\\=2.266\\$

<h3>since 2.266>1.645 we can reject the null hypothesis.</h3>