$5000 dollars is borrowed for 3 months at an annual interest rate of 18%. How much interest must be paid?

3. Jen’s parents buy a tablet for her to use in school. Of the $1500 cost, her parents expect her to pay back $1000 one year lat

NNADVOKAT [17]

Answer:

Step-by-step explanation:

Using the formula for calculating simple interest as shown;

Simple Interest = Principal * Rate *Time/100

Principal = Cost of tablet = $1500

Interest after one year = $1500-$1000 = $500

Time = 1year

Substituting this values into the formula;

$500 = 1500*rate*1/100\\Cross\ multiplying\\50,000 = 1500 * rate\\Rate = 50,000/1500\\Rate = 33.3%$

The interest rate that her parents assumed is 33.3%

4 0

2 years ago

Jerry was trying out for the cross country team. He has to be able to run an 8.5 min mile. If he ran 5 mi in 45 mins last weeken

lesantik [10]

Answer:no he misses the rate he needs to run by 30m seconds

Step-by-step explanation:

45/5=9

7 0

2 years ago

Read 2 more answers

A boy starts the week with 3kg of rice, every day he cooks 0.6kg. How much rice will he have left after 4 days?

almond37 [142]

Answer:

1.8 Kg

Step-by-step explanation:

a week- 3kg rice

every day-0.6 kg rice

therefore, multiply 3 kg by 0.6kg

3kg x 0.6kg =1.8 kg

he will have 1.8 kg rice after 4 days

have a great day !

5 0

1 year ago

5) In a certain supermarket, a sample of 60 customers who used a self-service checkout lane averaged 5.2 minutes of checkout tim

Ne4ueva [31]

Answer:

$\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643$

$S_p=2.940$

$t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751$

$df=60+72-2=130$

$p_v =P(t_{130}$

Assuming a significance level of $\alpha=0.05$ we have that the p value is lower than this significance level so then we can conclude that the mean for checkout time is significantly less for people who use the self-service lane

Step-by-step explanation:

Data given

Our notation on this case :

$n_1 =60$ represent the sample size for people who used a self service

$n_2 =72$ represent the sample size for people who used a cashier

$\bar X_1 =5.2$ represent the sample mean for people who used a self service

$\bar X_2 =6.1$ represent the sample mean people who used a cashier

$s_1=3.1$ represent the sample standard deviation for people who used a self service

$s_2=2.8$ represent the sample standard deviation for people who used a cashier

Assumptions

When we have two independent samples from two normal distributions with equal variances we are assuming that

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

The statistic is given by:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

And t follows a t distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

System of hypothesis

Null hypothesis: $\mu_1 \geq \mu_2$

Alternative hypothesis: $\mu_1 < \mu_2$

This system is equivalent to:

Null hypothesis: $\mu_1 - \mu_2 \geq 0$

Alternative hypothesis: $\mu_1 -\mu_2 < 0$

We can find the pooled variance:

$\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643$

And the deviation would be just the square root of the variance:

$S_p=2.940$

The statistic is given by:

$t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751$

The degrees of freedom are given by:

$df=60+72-2=130$

And now we can calculate the p value with:

$p_v =P(t_{130}$

Assuming a significance level of $\alpha=0.05$ we have that the p value is lower than this significance level so then we can conclude that the mean for checkout time is significantly less for people who use the self-service lane

5 0

2 years ago

Elton is a candle maker. Each 15 cm long candle he makes burns evenly for 6 hours if Elton makes a 45 cm long candle, how long w

tatuchka [14]

The candle would burn for 18 hours.

45 / 15 = 3

If the candle is three times longer than normal, it will burn three times longer than normal.

3 x 6 = 18

7 0

3 years ago