On the first day of a family road trip, Kai's family travels 365 miles.

If you’re using the inspection method, what constant term would you need in the numerator of this rational expression for (x+5)

Eva8 [605]

If I am using the inspection method, the constant term that I would need in the numerator of this rational expression for (x+5) to be a common factor of (x^2+6x+14) / (x+5) is 11/(x+1)

7 0

3 years ago

Read 2 more answers

Drag each tile to the correct box.

ohaa [14]

Answer:

Step-by-step explanation:

5 0

3 years ago

Harry invests £6000 in a savings account.

zysi [14]

Step-by-step explanation:

here,

principal (p)=£6000

Rate (R)=. 3.4%

time (t)=3 yrs

now,

total investment is compound amount i.e CA

here,

$CA = p(1 + \frac{r}{100} ) ^{t}$

=£6000(1+3.4/100)^3

=£6633.044

here he will get his total investment as £6633.044 in compound interest of 3.4%

7 0

2 years ago

A number line going from 0 to 4. Numbers 0, 1, 2, 3, and 4 are marked on the line. Marks divide the spaces between each number i

olga nikolaevna [1]

Answer:

8 pieces of string

Step-by-step explanation:

Given that :

Total yards of string = 4

Length in yd of each string = 1/2 yards

Number of pieces of string Ms. Cole has :

Total. Yards / length in yard of each string

4 / (1/2)

4 * (2 /1)

= 8/1

= 8 pieces of string.

6 0

3 years ago

National data indicates that 35% of households own a desktop computer. In a random sample of 570 households, 40% owned a deskt

Elan Coil [88]

Answer:

Yes, this provide enough evidence to show a difference in the proportion of households that own a desktop.

Step-by-step explanation:

We are given that National data indicates that 35% of households own a desktop computer.

In a random sample of 570 households, 40% owned a desktop computer.

Let p = population proportion of households who own a desktop computer

SO, Null Hypothesis, $H_0$ : p = 25% {means that 35% of households own a desktop computer}

Alternate Hypothesis, $H_A$ : p $\neq$ 25% {means that % of households who own a desktop computer is different from 35%}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of 570 households who owned a desktop computer = 40%

n = sample of households = 570

So, test statistics = $\frac{0.40-0.35}{{\sqrt{\frac{0.40(1-0.40)}{570} } } } }$

= 2.437

Since, in the question we are not given with the level of significance at which to test out hypothesis so we assume it to be 5%. Now at 5% significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics doesn't lies within the range of critical values of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that % of households who own a desktop computer is different from 35%.

3 0

3 years ago