I really need plz help ASAP the right answer will be marked of brainlest

Which of the following would correctly fill in the blank on the following statement? 1.7 x 10__ = 0.00017

katrin2010 [14]

The blank should be the power of -4, so the statement is 1.7 x 10^-4 = 0.00017.

8 0

3 years ago

Read 2 more answers

Please answer quick!!

Anon25 [30]

Answer:

if u answer my question i will answer your

Step-by-step explanation:

7 0

3 years ago

An architectural drawing lists the scale as 1/4" = 1'. if a bedroom measures 312" by 514" on the drawing, how large is the bedro

marusya05 [52]

The length of the bedroom exists at x = 9 and y = 6.

<h3>How to estimate the length of the bedroom?</h3>

From the given information, we get

Then $$\frac{(1/4)}{(9/4)} = \frac{1}{x}$

Solve this for x.

simplifying the value of x we get

Equate (1/9) to 1/x.

x = 9 (feet).

Convert 1.5 inches to feet using a proportion:

$$\frac{(1/4)}{(1.5)} = \frac{1}{y}$

Solve this for y.

simplifying the value of y we get

(1/4)y = 3/2

Multiply both sides of the equation by 4.

y = 6

Therefore, the length of the bedroom exists at x = 9 and y = 6.

To learn more the value of x refers to:

brainly.com/question/2284360

#SPJ4

6 0

1 year ago

The graph shows the pizza orders taken at a pizza shop during lunchtime one day.

miv72 [106K]

B cause it's true if you look at the graph

6 0

3 years 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

3 years ago