If x represents a number, then write an expression for three times the sum of x and 10

5X23X2 using properties to find the sum or product

lianna [129]

$5\cdot 23 \cdot 2=(5\cdot 2)\cdot 23 = 10\cdot 23=230$

4 0

3 years ago

Mr. Potter's class will take a test for 1 1/2 hours. They need to finish by 11:40 to get to lunch on time.What is the latest tim

stich3 [128]

The answer is 10:10

3 0

3 years ago

The cost c, to produce b baseball bats per day is modeled by the function c(b)=0.06b^2 - 7.2b + 390. What number of bats should

Finger [1]

You have two choices.
The minimum possible cost to operate your bat factory for a day is $390,
and there are two quantities of bats that both cost that much.

One possibility:
   Produce no bats at all per day. Zero. Nada. None.
   Cost = 0.06(0)² - 7.2(0) + 390 = $390 per day.

The other choice:
   Produce 120 bats per day.
   Cost = 0.06(120²) - 7.2(120) + 390
      = 0.06(14,400) - 7.2(120) + 390
= 864 - 864 + 390
= $390 per day.

If you produce any other number of bats in a day ... more than zero
but not 120 ... then it will cost you more than $390 to operate the factory
that day.

8 0

3 years ago

A public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes. Kar

ch4aika [34]

Answer:

We conclude that the mean waiting time is less than 10 minutes.

Step-by-step explanation:

We are given that a public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes.

Karen took bus number 14 during peak hours on 18 different occasions. Her mean waiting time was 7.8 minutes with a standard deviation of 2.5 minutes.

Let $\mu$ = mean waiting time for bus number 14.

So, Null Hypothesis, $H_0$ : $\mu \geq$ 10 minutes {means that the mean waiting time is more than or equal to 10 minutes}

Alternate Hypothesis, $H_A$ : $\mu$ < 10 minutes {means that the mean waiting time is less than 10 minutes}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean waiting time = 7.8 minutes

s = sample standard deviation = 2.5 minutes

n = sample of different occasions = 18

So, test statistics = $\frac{7.8-10}{\frac{2.5}{\sqrt{18} } }$ ~ $t_1_7$

= -3.734

The value of t test statistics is -3.734.

Now, at 0.01 significance level the t table gives critical value of -2.567 for left-tailed test.

Since our test statistic is less than the critical value of t as -3.734 < -2.567, 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 the mean waiting time is less than 10 minutes.

5 0

3 years ago

Need help with homework plz, also will give brainiest to who ever gets the question right first!

Degger [83]

Answer: x = 5/7

Step-by-step explanation:

The numerator is the power, while the denominator is the index (thing next to radical).

7 0

2 years ago