Fine the volume please

Use the graph below to answer How many dozens of cupcakes will be made after 3 hours?

aleksley [76]

Answer: 12 dozens of cupcakes will be made after three hours.

Step-by-step explanation: If you got to the three under hours and move up until the arrow intersects you will get 12 dozens

5 0

3 years ago

HELP giving brainliest: express the following ratio as a unit rate. Round to the nearest tenth, if necessary. 400 yards in 62.5

Elden [556K]

The unit rate is 6.4 yards per second.

Step-by-step explanation:

Given statement is;

400 yards in 62.5 seconds

As we have to find unit rate in terms of seconds; therefore, we will divide total yards by total seconds.

62.5 seconds = 400 yards

Dividing both sides by 62.5

$\frac{62.5}{62.5}\ seconds=\frac{400}{62.5}\ yards\\1\ second = 6.4\ yards$

The unit rate is 6.4 yards per second.

Keywords: unit rate, division

Learn more about unit rate at:

#LearnwithBrainly

4 0

4 years ago

Land's Bend sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the intern

Naya [18.7K]

Answer:

80% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases ( $\mu_1-\mu_2$ ) is [-9.132 , 23.332].

Step-by-step explanation:

We are given that a random sample of 7 sales receipts for mail-order sales results in a mean sale amount of $81.70 with a standard deviation of $18.75.

A random sample of 11 sales receipts for internet sales results in a mean sale amount of $74.60 with a standard deviation of $28.25.

Firstly, the Pivotal quantity for 80% confidence interval for the difference between population means is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t__n__1-_n__2-2$

where, $\bar X_1$ = sample mean sales receipts for mail-order sales = $81.70

$\bar X_2$ = sample mean sales receipts for internet sales = $74.60

$s_1$ = sample standard deviation for mail-order sales = $18.75

$s_2$ = sample standard deviation for internet sales = $28.25

$n_1$ = size of sales receipts for mail-order sales = 7

$n_2$ = size of sales receipts for internet sales = 11

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(7-1)\times 18.75^{2} +(11-1)\times 28.25^{2} }{7+11-2} }$ = 25.11

Here for constructing 80% confidence interval we have used Two-sample t test statistics as we don't know about population standard deviations.

So, 80% confidence interval for the difference between population means, ( $\mu_1-\mu_2$ ) is ;

P(-1.337 < $t_1_6$ < 1.337) = 0.80 {As the critical value of t at 16 degree

of freedom are -1.337 & 1.337 with P = 10%}

P(-1.337 < $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 1.337) = 0.80

P( $-1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ${(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}$ < $1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.80

P( $(\bar X_1-\bar X_2)-1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ( $\mu_1-\mu_2$ ) < $(\bar X_1-\bar X_2)+1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.80

80% confidence interval for ( $\mu_1-\mu_2$ ) =

[ $(\bar X_1-\bar X_2)-1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ , $(\bar X_1-\bar X_2)+1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ]

= [ $(81.70-74.60)-1.337 \times {25.11 \times \sqrt{\frac{1}{7} +\frac{1}{11} } }$ , $(81.70-74.60)+1.337 \times {25.11 \times \sqrt{\frac{1}{7} +\frac{1}{11} } }$ ]

= [-9.132 , 23.332]

Therefore, 80% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases ( $\mu_1-\mu_2$ ) is [-9.132 , 23.332].

4 0

3 years ago

A gallup survey indicated that 72% of 18- to 29-year-olds, if given choice, would prefer to start their own business rather than

Neko [114]

Answer:

The probability that no more than 70% would prefer to start their own business is 0.1423.

Step-by-step explanation:

We are given that a Gallup survey indicated that 72% of 18- to 29-year-olds, if given choice, would prefer to start their own business rather than work for someone else.

Let $\hat p$ = sample proportion of people who prefer to start their own business

The z-score probability distribution for the sample proportion is given by;

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

where, p = population proportion who would prefer to start their own business = 72%

n = sample of 18-29 year-olds = 600

Now, the probability that no more than 70% would prefer to start their own business is given by = P( $\hat p$ $\leq$ 70%)

P( $\hat p$ $\leq$ 70%) = P( $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ $\leq$ $\frac{0.70-0.72}{\sqrt{\frac{0.70(1-0.70)}{600} } }$ ) = P(Z $\leq$ -1.07) = 1 - P(Z < 1.07)

= 1 - 0.8577 = 0.1423

The above probability is calculated by looking at the value of x = 1.07 in the z table which has an area of 0.8577.

3 0

4 years ago

HURRY!!!! Determine the coordinates of the point on the unit circle corresponding to the given central angle. If necessary, roun

nadezda [96]

Answer:

The answer is c. (0,-1)

Step-by-step explanation:

So the unit circle is derived from a right triangle.

3 0

3 years ago